Some Basic Concepts of Chemistry

Density of HCl solution $(\mathrm{d})=1.13 \mathrm{~g} / \mathrm{ml}$

$ \mathrm{V}=300 \mathrm{ml} $

Wt. of HCl solution $= \mathrm{density} \times \mathrm{volume}=1.13 \times 300=339 \mathrm{~g}$

Wt. of $\mathrm{HCl}=339 \times \frac{38.55}{100}=130.68 \mathrm{~g}$ (LR)

$ \begin{aligned} & \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \\ & \text{Molar mass of } \mathrm{CaCO}_3 =40+12+(3 \times 16)=100 \mathrm{~g~mol^{-1}} \\ & \text{So, moles of } \mathrm{CaCO}_3=\frac{90}{100}=0.90 \text { mole } \\ & \text{Molar mass of } \mathrm{HCl}=1+35.5=36.5 \mathrm{~g~mol^{-1}} \\ & \text{So, moles of } \mathrm{HCl}=\frac{130.68}{36.5}=3.58 \text { mole } \\ & \text{From the equation, } 1 \text{ mol } \mathrm{CaCO}_3 \text{ needs } 2 \text{ mol } \mathrm{HCl} \\ & \text{So, } 0.90 \text{ mol } \mathrm{CaCO}_3 \text{ needs } 2 \times 0.90=1.80 \text{ mol } \mathrm{HCl} \end{aligned} $

Since available HCl $=3.58$ mol and required HCl $=1.80$ mol, HCl is in excess and $\mathrm{CaCO}_3$ is the limiting reagent (LR).

Moles of HCl remained $=3.58-1.80=1.78$ mole.

Mass of HCl remained $=1.78 \times 36.5=64.97 \mathrm{~g}$.

Primary standard (as per NCERT idea) should be a substance which is pure, stable in air, non-hygroscopic, readily soluble in water, and reacts completely with the titrant in a stoichiometric way.

Now check each statement:

A. Hydrated salts can be used as primary standard.

Some hydrated salts are sufficiently stable and can be used (example often cited: Mohr’s salt). So this can be correct.

B. Primary standard should not undergo any reaction with air.

Correct. It should not react with $O_2$, $CO_2$ or moisture of air.

C. Reactions of primary standard with another substance should be instantaneous and stoichiometric.

Correct. The reaction should be fast and have a definite stoichiometry so the end point is sharp.

D. Primary standard should not be soluble in water.

Incorrect. It should be readily soluble so that a standard solution can be prepared.

E. Primary standard should have low relative molar mass.

Incorrect. It should preferably have high molar mass to minimise weighing error.

So, correct statements are A, B and C only.

Correct option: Option D (A, B and C Only).

Balanced reaction:

$A + 2B \longrightarrow AB_2$

Step 1: Calculate moles of reactants

For $A$:

$n_A=\frac{36.0}{60}=0.60\ \text{mol}$

For $B$:

$n_B=\frac{56.0}{80}=0.70\ \text{mol}$

Step 2: Find the limiting reagent (NCERT method: compare required moles)

For $0.60$ mol of $A$, required moles of $B$:

$n_B(\text{required})=2 \times 0.60=1.20\ \text{mol}$

But available $B = 0.70$ mol, which is less than $1.20$ mol.

So, $B$ is the limiting reagent.

Hence statement A is false.

Step 3: Moles and mass of product formed

From the reaction, $2$ mol $B$ gives $1$ mol $AB_2$.

So moles of $AB_2$ formed:

$n_{AB_2}=\frac{0.70}{2}=0.35\ \text{mol}$

Molar mass of $AB_2$:

$M_{AB_2}=60 + 2(80)=220\ \text{g mol}^{-1}$

So statement C is false.

Mass of $AB_2$ formed:

$m_{AB_2}=0.35 \times 220=77.0\ \text{g}$

So statement B is true.

Step 4: Unreacted $A$

Moles of $A$ consumed = moles of $AB_2$ formed (ratio $1:1$):

$n_A(\text{consumed})=0.35\ \text{mol}$

Mass of $A$ consumed:

$m_A(\text{consumed})=0.35 \times 60=21.0\ \text{g}$

Unreacted $A$:

$m_A(\text{left})=36.0-21.0=15.0\ \text{g}$

So statement D is true.

Correct statements: B and D only

✅ Option D

$\begin{aligned} & 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})+6 \mathrm{Cl}^{-}(\mathrm{aq})+ 3 \mathrm{H}_2(\mathrm{g}) \\ & \text { From the balanced equation, }6\text{ moles of }\mathrm{HCl}\text{ give }3\text{ moles of }\mathrm{H}_2. \\ & \text { So, moles of }\mathrm{H}_2\text{ produced } \\ & =\frac{3}{6}\times \text{ moles of }\mathrm{HCl}\text{ used } \\ & =\frac{1}{2}\times \text{ moles of }\mathrm{HCl}\text{ used } \\ & \text { Also, }2\text{ moles of }\mathrm{Al}\text{ give }3\text{ moles of }\mathrm{H}_2. \\ & \text { So, moles of }\mathrm{H}_2\text{ produced } \\ & =\frac{3}{2}\times \text{ moles of }\mathrm{Al}\text{ used } \end{aligned}$

The chemical formula for magnesium pyrophosphate is $Mg_2P_2O_7$.

Using the given molar masses (Mg = 24, P = 31, O = 16):

$ \text{Molar mass of } Mg_2P_2O_7 = (2 \times 24) + (2 \times 31) + (7 \times 16) $

$ \text{Molar mass} = 48 + 62 + 112 = 222 \text{ g mol}^{-1} $

One mole of $Mg_2P_2O_7$ contains 2 moles of phosphorus atoms.

$ \text{Mass of P} = 2 \times 31 = 62 \text{ g} $

This means that 222 g of $Mg_2P_2O_7$ contains 62 g of phosphorus.

The mass of $Mg_2P_2O_7$ obtained in the experiment is 1.79 g.

Applying the unitary method:

$ \text{Mass of P in 1.79 g of } Mg_2P_2O_7 = \frac{62}{222} \times 1.79 \text{ g} $

$ \text{Mass of P} = \frac{110.98}{222} \approx 0.4999 \text{ g} $

The percentage of an element in a given compound is calculated as:

$ \% \text{ of P} = \frac{\text{Mass of P}}{\text{Total mass of compound (X)}} \times 100 $

$ \% \text{ of P} = \frac{0.4999 \text{ g}}{1.00 \text{ g}} \times 100 = 49.99\% $

Rounding off to the nearest integer, we get 50%.

Option D is correct.

Balance the reaction first (NCERT method):

$\text{MnO}_2(s) + 4\text{HCl}(aq) \rightarrow \text{MnCl}_2(aq) + \text{Cl}_2(g) + 2\text{H}_2\text{O}(l)$

Step 1: Calculate moles of $\text{MnO}_2$

Molar mass of $\text{MnO}_2$:

$M(\text{MnO}_2)=55 + 2(16)=55+32=87\ \text{g mol}^{-1}$

Moles of $\text{MnO}_2$ in $8.7\ \text{g}$:

$n(\text{MnO}_2)=\frac{8.7}{87}=0.1\ \text{mol}$

Step 2: Use mole ratio to find moles of $\text{Cl}_2$

From the balanced equation:

$1\ \text{mol MnO}_2 \rightarrow 1\ \text{mol Cl}_2$

So,

$n(\text{Cl}_2)=0.1\ \text{mol}$

Step 3: Convert moles of $\text{Cl}_2$ to mass

Molar mass of $\text{Cl}_2$:

$M(\text{Cl}_2)=2\times 35.5=71\ \text{g mol}^{-1}$

Mass of $\text{Cl}_2$:

$m(\text{Cl}_2)=n\times M = 0.1 \times 71 = 7.1\ \text{g}$

Answer: $7.1\ \text{g}$ (Option C)

Balanced reaction (NCERT method):

$\mathrm{Ca} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{H_2}$

Step 1: Moles of calcium

Given mass of Ca $=14.0\ \mathrm{g}$, molar mass of Ca $=40\ \mathrm{g\,mol^{-1}}$

$n(\mathrm{Ca})=\frac{14.0}{40}=0.35\ \mathrm{mol}$

Since $\mathrm{HCl}$ is in excess, the limiting reagent is Ca.

So Option A is correct.

Step 2: Amount of products formed (from stoichiometry)

From the balanced equation:

• $1\ \mathrm{mol\ Ca} \rightarrow 1\ \mathrm{mol\ CaCl_2}$

• $1\ \mathrm{mol\ Ca} \rightarrow 1\ \mathrm{mol\ H_2}$

Hence,

$n(\mathrm{CaCl_2})=0.35\ \mathrm{mol}$

$n(\mathrm{H_2})=0.35\ \mathrm{mol}$

So Option D is correct.

Step 3: Volume of $\mathrm{H_2}$ at $1\ \mathrm{atm}$ and $273\ \mathrm{K}$ (STP)

At STP, $1\ \mathrm{mol}$ gas occupies $22.4\ \mathrm{L}$.

$V(\mathrm{H_2})=0.35\times 22.4=7.84\ \mathrm{L}$

So Option C is correct.

Step 4: Mass of $\mathrm{CaCl_2}$

Molar mass of $\mathrm{CaCl_2}$:

$=40 + 2(35.5)=40+71=111\ \mathrm{g\,mol^{-1}}$

Mass produced:

$m(\mathrm{CaCl_2})=0.35\times 111=38.85\ \mathrm{g}$

But Option B says $33.3\ \mathrm{g}$, which is wrong.

Incorrect statement: Option B.

In the combustion of organic compound, all "C" in $\mathrm{CO}_2$ and all " H " in $\mathrm{H}_2 \mathrm{O}$ comes from organic compound

Weight of " C " in $\mathrm{CO}_2=\frac{12}{44} \times 0.307$

$=0.0837 \mathrm{~gm}$

Weight of " H " in $\mathrm{H}_2 \mathrm{O}=\frac{2}{18} \times 0.127=0.0141 \mathrm{~g}$

$\begin{aligned} &\begin{gathered} \% ' H^{\prime} \text { in compound }=\frac{0.0141}{0.21} \times 100=6.719 \% \\ =6.72 \% \end{gathered}\\ &\text { Weight of "O" in compound }\\ &\begin{aligned} & =0.210-(0.0837+0.0141) \\ & =0.1122 \end{aligned}\\ &\begin{aligned} & \% \text { of "O" in compound }=\frac{0.1122}{0.21} \times 100 \\ & =53.41 \% \end{aligned} \end{aligned}$

We are tasked with calculating the mass of magnesium required to produce 220 mL of hydrogen gas at STP with excess dilute HCl.

Step 1: Write the balanced reaction

$ \text{Mg} + 2 \text{HCl} \;\;\rightarrow\;\; \text{MgCl}_2 + \text{H}_2 $

• 1 mole Mg gives 1 mole H₂.

Step 2: Molar volume at STP

At STP, 1 mole of an ideal gas occupies 22.4 L = 22,400 mL.

Step 3: Moles of H₂ produced

$ n(\text{H}_2) = \frac{\text{Volume of H₂}}{\text{Molar Volume}} = \frac{220}{22400} = 0.00982 \,\text{mol} $

Step 4: Moles of Mg required

From stoichiometry, 1 mole Mg produces 1 mole H₂, so:

$ n(\text{Mg}) = n(\text{H}_2) = 0.00982 \,\text{mol} $

Step 5: Mass of Mg required

$ m(\text{Mg}) = n \times M(\text{Mg}) = 0.00982 \times 24 = 0.236 \,\text{g} $

$ = 236 \,\text{mg} $

✅ Final Answer:

Option D: 236 mg

When $10 \mathrm{ml}, 2 \mathrm{M} \mathrm{NaOH}$ solution is added to 20 ml of 1 M HCl solution :

$\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$

$\begin{array}{rlrl} \text { Initial : MV } & =2 \times 0.1 & & M V=1 \times 0.2 \\ & =0.2 \text { mole } & & =0.2 \mathrm{~mole} \\ \text { Final } & 0 & 0 \end{array}$

$\therefore$ Resulting solution becomes neutral.

Now when 10 mol of above solution is poured into a flask containing 2 mole HCl and made solution 100 ml will distilled water.

Molarity of $\mathrm{HCl}=\frac{2}{100} \times 1000=20$

Step 1: Calculate the pressure of dry nitrogen gas

The nitrogen gas was collected over water, so we subtract the aqueous tension (pressure of water vapor) from the total pressure to find the pressure of dry nitrogen gas.

$\text{Pressure of dry } \mathrm{N}_2 = 715\,\mathrm{mm\,Hg} - 15\,\mathrm{mm\,Hg} = 700\,\mathrm{mm\,Hg}$

Now, convert this pressure from mm Hg to atm:

$700\,\mathrm{mm\,Hg} \div 760 = \frac{700}{760}\,\mathrm{atm}$

Step 2: Use the ideal gas equation to find moles of nitrogen

The volume of nitrogen ($V$) is given as 60 mL. Convert this to liters: 60 mL = 0.060 L (or use $60 \times 10^{-3}$ L). The temperature ($T$) is 300 K, and the gas constant ($R$) is 0.0821 L·atm·K⁻¹·mol⁻¹.

Using the formula $\mathrm{n} = \frac{\mathrm{PV}}{\mathrm{RT}}$:

$ \mathrm{n} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} $

When you calculate this, you get: $ \mathrm{n} = 0.0022\;\text{mole} $

Step 3: Calculate the mass of nitrogen gas produced

Moles of nitrogen are 0.0022. The molar mass of $\mathrm{N}_2$ is 28 g/mol.

$ \text{Mass of } \mathrm{N}_2 = 0.0022 \times 28 = 0.063\,\mathrm{g} $

Step 4: Find the percentage of nitrogen in the compound

The mass of organic compound taken is 0.4 g. To find the percentage of nitrogen:

$ \text{Percentage of nitrogen} = \frac{\text{Mass of } \mathrm{N}_2}{\text{Mass of compound}} \times 100 $

$ = \frac{0.063}{0.4} \times 100 = 15.71\% $

To determine which element has the highest number of atoms among 10^-9 grams of the elements Pb, Po, Pr, and Pt, we use the formula for calculating the number of atoms:

$ \text{Number of atoms} = \frac{\text{Mass}}{\text{Molar Mass (g/mol)}} \times N_A $

where $ N_A $ is Avogadro's number.

From this formula, it's clear that for a given mass, the element with the smallest molar mass will have the greatest number of atoms, as it appears in the denominator.

Below are the molar masses of the elements:

Molar mass of Pr (Praseodymium), $ M_{\text{Pr}} = 141 $ g/mol

Molar mass of Pt (Platinum), $ M_{\text{Pt}} = 195 $ g/mol

Molar mass of Pb (Lead), $ M_{\text{Pb}} = 207 $ g/mol

Molar mass of Po (Polonium), $ M_{\text{Po}} = 209 $ g/mol

Since Praseodymium (Pr) has the smallest molar mass ($ M_{\text{Pr}} = 141 $ g/mol), 10^-9 grams of Pr will contain the highest number of atoms compared to the other elements listed.

The reaction is:

$ \mathrm{CaCO}_3(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) $

Calculate Moles of $\mathrm{CaCO}_3$:

The molar mass of $\mathrm{CaCO}_3$ is calculated as

$ 40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol} $

Given 1000 g of $\mathrm{CaCO}_3$, the number of moles is

$ \frac{1000}{100} = 10 \ \mathrm{moles} $

Calculate Moles of $\mathrm{HCl}$:

Using the given concentration and volume for $\mathrm{HCl}$, the moles are calculated as follows:

$ 0.76 \times \frac{250}{1000} = 0.19 \ \mathrm{moles} $

$\mathrm{HCl}$ is the limiting reagent (L.R.) because it has fewer moles than needed to completely react with the $\mathrm{CaCO}_3$.

Calculate Moles of $\mathrm{CaCl}_2$ Formed:

According to the stoichiometry of the reaction, 2 moles of $\mathrm{HCl}$ produce 1 mole of $\mathrm{CaCl}_2$. Therefore, the moles of $\mathrm{CaCl}_2$ formed are:

$ \frac{0.19}{2} = 0.095 \ \mathrm{moles} $

Calculate Mass of $\mathrm{CaCl}_2$:

The molar mass of $\mathrm{CaCl}_2$ is:

$ 40 + 2 \times 35.5 = 111 \ \mathrm{g/mol} $

Thus, the mass of $\mathrm{CaCl}_2$ is:

$ 0.095 \times 111 = 10.545 \ \mathrm{g} $

Hence, the mass of $\mathrm{CaCl}_2$ formed is 10.545 g.

Calculate the Moles of Carbon:

Carbon is fully converted to $ \mathrm{CO}_2 $.

Moles of $ \mathrm{CO}_2 = \frac{1.46\, \text{g}}{44\, \text{g/mol}} = 0.033 \text{ mol} $.

Moles of Carbon ($ \text{C} $) = Moles of $ \mathrm{CO}_2 $ = 0.033 mol.

Mass of Carbon = $ 0.033 \times 12\, \text{g/mol} = 0.396\, \text{g} $.

Calculate the Moles of Hydrogen:

Hydrogen is fully converted to $ \mathrm{H}_2\mathrm{O} $.

Moles of $ \mathrm{H}_2\mathrm{O} = \frac{0.567\, \text{g}}{18\, \text{g/mol}} = 0.0315 \text{ mol} $.

Moles of Hydrogen ($ \text{H} $) = $ 2 \times 0.0315 = 0.063 \text{ mol} $.

Mass of Hydrogen = $ 0.063 \times 1\, \text{g/mol} = 0.063\, \text{g} $.

Calculate the Mass of Oxygen:

Total mass of compound $ \mathrm{X} $ = 1.0 g.

Mass of Oxygen = Total mass - (Mass of Carbon + Mass of Hydrogen)

Mass of Oxygen = $ 1.0\, \text{g} - (0.396\, \text{g} + 0.063\, \text{g}) = 0.541\, \text{g} $.

Moles of Oxygen ($ \text{O} $) = $ \frac{0.541}{16} = 0.0338 \approx 0.033 \text{ mol} $.

Determine the Empirical Formula:

The ratio of moles: $ \text{C} = 0.033 $, $ \text{H} = 0.063 $, $ \text{O} = 0.033 $.

Simplified, this ratio is approximately 1:2:1.

Empirical formula: $ \mathrm{CH}_2\mathrm{O} $.

Calculate the Empirical Formula Mass:

Empirical formula mass = $ 12 \times 1 + 1 \times 2 + 16 \times 1 = 30\, \text{g/mol} $.

Therefore, the empirical formula mass of compound $ \mathrm{X} $ is 30 g/mol.

A) Weight of a substance is the amount of matter present in it.

Not correct.

The amount of matter present in a substance is called its mass, not its weight.

Mass of a substance is the amount of matter present in it.

B) Mass is the force exerted by gravity on an object.

Not correct.

Mass is not the force exerted on an object by gravity. Mass is the amount of matter in an object, while weight is the force of gravity on that object.

C) Volume is the amount of space occupied by a substance.

Correct.

The amount of space occupied by a substance is its volume, which is simply the equivalent three dimensional measurements of spatial geometry (height, width; and (length).

Standard units of volume are cubic centimeters for solids, milliliters or liters for liquids, and liters pr gases.

D) Temperatures below $0^{\circ} \mathrm{C}$ are possible in Celsius scale, but in Kelvin scale negative temperature is not possible.

Correct.

Celsius temp is possible below $0^{\circ} \mathrm{C}$ (negative temp). But, it is not possible in the Kelvin Scale

$273 \mathrm{~K} \rightarrow$ Absolute temperature

Kelvin temp $={ }^{\circ} \mathrm{C}$ temp +273.15

The minimum temperature possible in the Kelvin scale is the absolute zero temperature ($\mathrm{OK}$). OK means $-273.15^{\circ} \mathrm{C}$. Since absolute zero is the coldest possible temperature, there we no negative values on the Kelvin temperature scale.

E) Precision refers to the closeness of various measurements for the same quantity.

Correct.

A measure of how close repeated measurements are to each other is known as precision. Precision is the reproducibility of a result or measurement.

If weight of a substance is measured five times and get 8.2 kg each time, then the measurement is very precise - Example.

So, the correct statements are (C) , (D) and (E).

Correct answer: Option 3) (C) ,(D) and(E) only

$\% \mathrm{w} / \mathrm{w}$ of $\mathrm{HNO}_3=75 \%$

means 100 gm of solution containing 75 g of $\mathrm{HNO}_3$

$\&\left(\frac{\mathrm{gm}}{\mathrm{m}_1}\right)_{\text {solution }}=1.25=\frac{100 \mathrm{gm}}{\mathrm{V}}$

$\mathrm{V}_{\mathrm{ml}}$ of 100 gm solution $=\frac{100}{1.25} \mathrm{ml}$

$\because 75 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in $\frac{100}{1.25} \mathrm{ml}$ solution

$\therefore 30 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in

$\frac{100}{1.25 \times 75} \times 30=32 \mathrm{ml} \text { solution }$

$\begin{array}{lllll} \mathrm{C} & : & \mathrm{H} & : & \mathrm{O} \\ \frac{54.2}{12} & : & 9.2 & : & \frac{36.6}{16} \\ 4.516 & : & 9.2 & : & 2.287 \\ \frac{4.516}{2.287} & : & \frac{9.2}{2.287} & : & \frac{2.287}{2.287} \\ 1.97 & : & 4.02 & : & 1 \end{array}$

$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \Rightarrow$ Empirical formula

E.F. mass $=24+4+16=44$

and molar mass $=132$

$\begin{aligned} \text { Hence molecular formula } & =\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_3 \\ & =\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \end{aligned}$

Correct Option (3)

$\begin{aligned} & (\text { moles })_{\text {initial }}=\frac{x \times 10^{-3}}{44} \\ & (\text { moles })_{\text {removal }}=\frac{10^{21}}{6.02 \times 10^{23}} \\ & (\text { moles })_{\text {left }}=(\text { moles })_{\text {initial }}-(\text { moles })_{\text {removed }} \\ & 2.8 \times 10^{-3}=\frac{x \times 10^{-3}}{44}-\frac{10^{21}}{6.02 \times 10^{23}} \\ & \Rightarrow x=196.2 \mathrm{mg} \end{aligned}$

$ \text{Molarity of NaCl} = 3 \text{ M} \quad (\text{3 moles in 1 L of solution}) $

Given the density of the solution is $1.25 \, \text{g/mL},$ the total mass of 1 liter (1000 mL) of the solution is:

$ 1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g} $

To find the molality, we need the mass of the solvent (water). First, calculate the mass of NaCl in 1 L of solution. The molar mass of NaCl is approximately:

$ 23 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} \approx 58.45 \, \text{g/mol} $

Thus, the mass of NaCl is:

$ 3 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 175.35 \, \text{g} $

Now, the mass of the solvent (water) is:

$ 1250 \, \text{g (solution)} - 175.35 \, \text{g (NaCl)} \approx 1074.65 \, \text{g} $

Convert the mass of the solvent to kilograms:

$ 1074.65 \, \text{g} \div 1000 \approx 1.07465 \, \text{kg} $

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent:

$ m = \frac{3 \, \text{mol}}{1.07465 \, \text{kg}} \approx 2.79 \, \text{mol/kg} $

Thus, the molality of the solution is approximately $2.79 \, \text{m}.$

Step 1: Find the mass of sodium (Na) in the mixture.

Na makes up 30% of the total mixture.

So, mass of Na = $0.30 \times 4 = 1.2$ g

Step 2: Set up variables for the unknowns.

Let $x$ = mass of NaCl in grams.
Let $y$ = mass of NaBr in grams.

Step 3: Write the first equation (total mass).

$x + y = 4 \quad ...(i)$

Step 4: Write the second equation (mass of Na from both salts).

Molar mass of NaCl = $23 + 35.5 = 58.5$ u.
Molar mass of NaBr = $23 + 80 = 103$ u.

Each gram of NaCl contains $\frac{23}{58.5}$ g of Na, and each gram of NaBr contains $\frac{23}{103}$ g of Na.

So, total mass of Na from both salts is:

$x \times \frac{23}{58.5} + y \times \frac{23}{103} = 1.2 \quad ...(ii)$

Step 5: Solve the two equations.

From (i): $y = 4 - x$

Substitute $y$ into (ii):

$x \times \frac{23}{58.5} + (4 - x) \times \frac{23}{103} = 1.2$

(Solving this equation for $x$ gives $x = 1.8068$ g.)

Step 6: Find the percentage of NaCl in the mixture.

$\% \text{ of NaCl} = \frac{\text{Mass of NaCl}}{\text{Total mass}} \times 100 = \frac{1.8068}{4} \times 100 \approx 45\%$

Normality of $\mathrm{Na}_2 \mathrm{CO}_3=$ molarity $\times$n$-factor

$ =0.5 \mathrm{M} \times 2=1 \mathrm{~N} $

Normality of acid $A$,

$ \begin{aligned} & N_{\text {acid } A} \times V_{\text {acid } A}=N_{\mathrm{Na}_2 \mathrm{CO}_3} \times V_{\mathrm{Na}_2 \mathrm{CO}_3} \\ & N_{\text {acid } A}=\frac{1 \times 25}{10}=2.5 \mathrm{~N} \end{aligned} $

Similarly normality of acid $B$

$ \mathrm{N}_{\text {acid } B}=0.625 \mathrm{~N} $

Volume of acid $A$ needed for 1 L of 1 N solution

$ \begin{aligned} & N_{\text {acid } A} \times V_{A \text { final }}=N_{\text {final }} \times V_{\text {final }} \\ & V_{A \text { final }}=\frac{1 \times 1}{2.5}=0.4 \mathrm{~L} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Volume of acid $B$ needed for 1 L of 1 N solution,

$ V_{B \text { final }}=\frac{1 \times 1}{0.625}=1.6 \mathrm{~L} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Total moles of gas,

$ \begin{aligned} n_{\text {total }} & =\frac{\text { Volume }}{\text { Molar volume at STP }} \\ & =\frac{28}{224}=1.25 \mathrm{moles} \end{aligned} $

$ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_6+\frac{7}{2} \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Moles of $\mathrm{O}_2=\frac{128}{32}=4 \mathrm{~mol}$

Let $x$ be the moles of $\mathrm{C}_2 \mathrm{H}_6$ and $y$ be moles of $\mathrm{C}_2 \mathrm{H}_4$

$ \begin{aligned} & x+y=1.25 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{7}{2} x+3 y=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solve for $y, y=0.75 \mathrm{~mol}$

Mole fraction of $\mathrm{C}_2 \mathrm{H}_4=\frac{0.75}{1.25}=0.6$

The balanced equation for complete reaction is,

$ 2 \mathrm{C}_2 \mathrm{H}_6+7 \mathrm{O}_2 \longrightarrow 4 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} $

Moles of $\mathrm{C}_2 \mathrm{H}_6=\frac{15 \mathrm{~g}}{30.07}=0.499 \mathrm{~mol}$

Moles of $\mathrm{O}_2=\frac{112}{32}=3.50 \mathrm{~mol}$

Moles of $\mathrm{O}_2$ required for 0.499 mol of $\mathrm{C}_2 \mathrm{H}_6$

$ 0.499 \times \frac{7}{2}=1.747 \mathrm{~mol} \mathrm{O}_2 $

Since, $3.50 \mathrm{~mol}, \mathrm{O}_2$ is available and only 1.747 mol .

$\mathrm{O}_2$ is needed, $\mathrm{C}_2 \mathrm{H}_6$ is limiting reactant. Mol of $\mathrm{CO}_2$ produced

$ 0.499 \times \frac{4}{2}=0.998 \mathrm{~mol} \mathrm{CO}_2 $

mol of $\mathrm{H}_2 \mathrm{O}$ produced

$ 0.499 \times \frac{6}{2}=1.497 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} $

Moles of excess $\mathrm{O}_2=3.50-1.747$

$ =1.753 \mathrm{~mol} $

Total mol of gaseous substance, $0.998+1.497+1.753=4.25 \mathrm{~mol}$

Let $x$ and $y$ be number of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ ions.

$ x+y=0.93 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Total negative charge from oxygen is,

$ 1 \times 2=2 $

Total positive charge from Fe ions

$ =2 x+3 y $

For charge neutrality $=2 x+3 y=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii)

$ y=0.14 $

$ \begin{aligned} & \mathrm{Fe}^{3+} \%=\frac{\text { Number of } \mathrm{Fe}^{3+} \text { ions }}{\text { Total ions }} \times 100 \\ \Rightarrow & \frac{0.14}{0.93} \times 100 \approx 15 \% \end{aligned} $

Let's find the number of significant figures carefully.

(A) $ 0.0025 $

• Leading zeros are not significant.

• Only the digits 2 and 5 are significant.

✅ Significant figures = 2

(B) $ 500.0 $

• Trailing zeros after the decimal point are significant.

• $ 500.0 $ → digits 5, 0, 0, and 0 after decimal.

✅ Significant figures = 4

(C) $ 2.0034 $

• All nonzero digits are significant.

• Zeros between nonzero digits are significant.

• So digits: 2, 0, 0, 3, 4 → all count.

✅ Significant figures = 5

✅ Final Answer

(A, B, C) ⇒ (2, 4, 5)

Correct Option: D

$ \boxed{2,\,4,\,5} $

Let $x$ be the mass of $\mathrm{CaCO}_3$,y be the mass of $\mathrm{MgCO}_3$

$ x+y=1.84 \mathrm{~g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When $\mathrm{CaCO}_3$ deompose

$ \underset{100 \mathrm{~g}}{\mathrm{CaCO}_3} \longrightarrow \underset{56 \mathrm{~g}}{\mathrm{CaO}}+\mathrm{CO}_2 $

Mass of CaO produced $=\frac{56}{100} x$

$ \underset{84}{\mathrm{MgCO}_3} \longrightarrow \underset{40}{\mathrm{MgO}}+\mathrm{CO}_2 $

Mass of MgO produced $=\frac{40}{84} x$

Total mass of residue

$ \frac{56}{100} x+\frac{40}{84} y=0.96 \mathrm{~g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii),

$ x=1 $

Mass of $\mathrm{CaCO}_3=1 \mathrm{~g}$

$ \% \text { of } \mathrm{CaCO}_3=\frac{1}{1.84} \times 100=54.34 \% $

Mass of chlorine is the difference between the mass of the chloride and mass of the elements.

$ 315.5-209=106.59 $

Equivalent weight of element

$ \begin{aligned} & =\frac{\text { Mass of element }}{\text { Mass of chlorine }} \times \begin{array}{l} \text { Atomic weight } \\ \text { of chlorine } \end{array} \\ & =\frac{209 \mathrm{~g}}{106.5 \mathrm{~g}} \times 35.5 \mathrm{u} \approx 69.66 \mathrm{u} \end{aligned} $

$\therefore$ Mass of oxygen

$ \begin{aligned} & =\frac{\text { Mass of element }}{\text { Equivalent weight }} \times \begin{array}{l} \text { Atomic weight } \\ \text { of oxygen } \end{array} \\ & \text { of element } \\ & =\frac{418 \mathrm{~g}}{69.66 \mathrm{u}} \times 16 \mathrm{u}=6 \times 16 \mathrm{~g}=96 \mathrm{~g} \end{aligned} $

The number of significant figures are

$ 0.0063=2 $

[Zero before number is non-significant]

$ 132.00=5 $

[Zero after decimals are significant]

$ 1004=4 $

[Zero between two non-zero are significant]

To determine the amount of oxygen required for the complete combustion of 900 g of glucose, we need to follow these steps:

First, let's write the balanced chemical equation for the combustion of glucose $(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)$:

$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 + 6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 + 6 \mathrm{H}_2 \mathrm{O}$

From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen $(\mathrm{O}_2)$.

Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as $180 \ \mathrm{g} \ \mathrm{mol}^{-1}$.

Number of moles of glucose = $\frac{900 \ \mathrm{g}}{180 \ \mathrm{g} \ \mathrm{mol}^{-1}} = 5 \ \mathrm{mol}$

According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen. Therefore, 5 moles of glucose will require:

Number of moles of oxygen = $5 \ \mathrm{mol} \times 6 = 30 \ \mathrm{mol}$

Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen $\left(\mathrm{O}_2\right)$ is $32 \ \mathrm{g} \ \mathrm{mol}^{-1}$.

Mass of oxygen = $30 \ \mathrm{mol} \times 32 \ \mathrm{g} \ \mathrm{mol}^{-1} = 960 \ \mathrm{g}$

Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is 960 g.

So, the correct option is Option D: 960.

To calculate the molality $(\mathrm{m})$ of a 3M (molar) aqueous solution of $\mathrm{NaCl}$, we need to follow certain steps, using the given data and understanding the definitions properly.

Molarity (M) is defined as the number of moles of solute per liter of solution. Molality $(\mathrm{m})$, on the other hand, is defined as the number of moles of solute per kilogram of solvent. The given density of the solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$ which allows us to calculate the mass of the solvent in a given volume of solution.

We know:

• Molarity (M) of NaCl = $3 \mathrm{~M}$
• Density of NaCl solution = $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$
• Molar mass of $\mathrm{NaCl} = \mathrm{Na} + \mathrm{Cl} = 23 + 35.5 = 58.5 \mathrm{~g} \mathrm{~mol}^{-1}$

First, to find the molality, we need to find the moles of NaCl in a given mass of the solvent. Keeping in mind that $3 \mathrm{~M}$ means there are $3 \mathrm{~moles}$ of $\mathrm{NaCl}$ per liter of solution, we consider 1 liter (or 1000 mL) of solution to simplify our calculation, knowing that density can convert volume to mass directly.

The mass of 1 liter of NaCl solution is $1.25 \mathrm{~g} \mathrm{~mL}^{-1} \times 1000 \mathrm{~mL} = 1250 \mathrm{~g}$.

Since we have a 3 M solution of NaCl,

$3 \mathrm{~moles} = \frac{\text{mass of the solution} - \text{mass of solvent}}{\text{molar mass of NaCl}}$

With 3 moles of NaCl, the mass of NaCl present in 1 L solution $= 3 \mathrm{~moles} \times 58.5 \mathrm{~g/mol} = 175.5 \mathrm{~g}$

The mass of the solvent (water) can be found by subtracting the mass of NaCl from the total mass of the solution.

$\text{Mass of solvent} = 1250 \mathrm{~g} - 175.5 \mathrm{~g} = 1074.5 \mathrm{~g}$

To convert this mass into kilograms (since molality is expressed per kilogram of solvent),

$1074.5 \mathrm{~g} = 1.0745 \mathrm{~kg}$

Finally, the molality $(\mathrm{m})$,

$\mathrm{m} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{3 \mathrm{~moles}}{1.0745 \mathrm{~kg}}$

$\mathrm{m} \approx 2.79$

So, the correct answer is

Option D: 2.79 m

To find the molarity, $x$, of the $\mathrm{NaOH}$ solution, we need to relate the given density and molality. Molarity is defined as the number of moles of solute per liter of solution, while molality is defined as the number of moles of solute per kilogram of solvent. We are given the following:

• Molality (m) = $3\ \mathrm{mol/kg}$ (3 molal)
• Density of solution ($\rho$) = $1.12\ \mathrm{g/mL}$ = $1120\ \mathrm{g/L}$
• Molar mass of $\mathrm{NaOH}$ ($M_\mathrm{NaOH}$) = $40\ \mathrm{g/mol}$

First, let's find the number of moles of $\mathrm{NaOH}$ in 1 kg of water, which is the definition of molality. Since the molality is 3 molal, we have:

$3\ \mathrm{mol}$ of $\mathrm{NaOH}$ in 1 kg of water (or 1000 g of water).

Now, to find the molarity, we need the volume of the solution in which these moles of $\mathrm{NaOH}$ are dissolved. The mass of 1 L (1000 mL) of solution, given the density, is:

$1120\ \mathrm{g}$

The mass of $\mathrm{NaOH}$ in 3 moles is:

$3 \mathrm{mol} \times 40\ \mathrm{g/mol} = 120\ \mathrm{g}$

Thus, the mass of water in this 1 L solution can be found by subtracting the mass of $\mathrm{NaOH}$ from the total mass of the solution:

$1120\ \mathrm{g} - 120\ \mathrm{g} = 1000\ \mathrm{g}$

So, we have 3 moles of $\mathrm{NaOH}$ in 1 L of solution. Therefore, the molarity (x) of the solution is directly 3 M (since molarity is moles of solute per liter of solution). Hence, $x = 3$.

Therefore, the correct answer is Option B: 3.0.

To find the number of moles of methane ($CH_4$) required to produce 11g of $CO_2$ upon complete combustion, we first need the balanced chemical equation for the combustion of methane:

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

From the balanced equation, we see that 1 mole of $CH_4$ produces 1 mole of $CO_2$. Therefore, the moles of $CH_4$ needed to produce a certain amount of $CO_2$ will be equal to the moles of $CO_2$ produced.

To find the moles of $CO_2$ produced from 11g of $CO_2$, we use the formula:

$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of $CO_2$ is $44 \mathrm{~g/mol}$ (12 from carbon and 16$\times$2 from oxygen). Then, the moles of $CO_2$ produced from 11g of $CO_2$ can be calculated as:

$\text{Moles of } CO_2 = \frac{11 \mathrm{~g}}{44 \mathrm{g/mol}} = 0.25 \mathrm{~mol}$

Since the stoichiometry of the reaction between methane and oxygen is 1:1 for $CH_4$ and $CO_2$, the moles of $CH_4$ required to produce 11g of $CO_2$ is also 0.25 moles.

Therefore, the correct answer is Option B: 0.25.

To determine the molecular formula of the compound, we first calculate the empirical formula based on the given percentages of carbon, hydrogen, and oxygen. Then we use the molecular weight to find the molecular formula.

Step 1: Calculating the moles of each element per 100 g of the compound

• For carbon (%C = 42.1%), the moles of carbon are calculated as follows:

  $\text{Moles of C} = \frac{\% \text{ by weight of C}}{\text{Atomic weight of C}} = \frac{42.1 \text{ g}}{12.01 \text{ g/mol}}$

• For hydrogen (%H = 6.4%), the moles of hydrogen are:

  $\text{Moles of H} = \frac{\% \text{ by weight of H}}{\text{Atomic weight of H}} = \frac{6.4 \text{ g}}{1.008 \text{ g/mol}}$

• For oxygen, since it makes up the remainder, we find its percentage by subtracting the percentages of carbon and hydrogen from 100%:

  $\% \text{O} = 100 - 42.1 - 6.4 = 51.5\%$

  Then the moles of oxygen are:

  $\text{Moles of O} = \frac{\% \text{ by weight of O}}{\text{Atomic weight of O}} = \frac{51.5 \text{ g}}{16.00 \text{ g/mol}}$

Step 2: Dividing the calculated moles by the smallest number of moles to find the simplest whole number ratio:

This involves dividing each mole value by the smallest of the mole values obtained. However, since we didn't calculate the exact number of moles for each element in the above step, let's proceed conceptually based on given percentages:

To find the simplest whole number ratio, we would perform the following calculations:

$\text{Moles of C} \approx \frac{42.1}{12}$

$\text{Moles of H} \approx \frac{6.4}{1}$

$\text{Moles of O} \approx \frac{51.5}{16}$

Step 3: Determining the empirical formula:

Given the lack of exact numbers from the prior step, let's assume each mole value is divided by the smallest mole value among the C, H, and O calculations. This ratio generally defines the empirical formula, which constitutes the smallest whole number ratio of atoms in the compound.

Step 4: Using the molecular weight to find the molecular formula:

Once we have the empirical formula, we would calculate its mass and compare it to the molecular weight provided (342). The molecular formula is a multiple of the empirical formula that matches the given molecular weight. Since we haven't calculated the exact empirical formula mass, let's examine the options directly considering a molecular weight of 342:

Option A: $C_{14}H_{20}O_{10}$ - The molar mass would be $14 \times 12 + 20 \times 1 + 10 \times 16 = 168 + 20 + 160 = 348 \text{, which is close but not 342.}$

Option B: $C_{12}H_{22}O_{11}$ - The molar mass would be $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \text{, which matches the given molecular weight.}$

Thus, the correct molecular formula based on the molecular weight and composition given would be Option B: $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$.

To find the molarity of the aqueous solution, we need to calculate the number of moles of NaCl and then divide it by the volume of the solution in liters.

First, determine the molar mass of NaCl:

$ \text{Molar mass of NaCl} = \text{Molar mass of Na} + \text{Molar mass of Cl} = 23 + 35.5 = 58.5 \, \text{g/mol} $

Next, calculate the moles of NaCl:

$ \text{Moles of NaCl} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol} $

Now, convert the volume of the solution from milliliters to liters:

$ 500 \, \text{mL} = 0.500 \, \text{L} $

Finally, calculate the molarity (M), which is the number of moles of solute per liter of solution:

$ M = \frac{\text{Moles of NaCl}}{\text{Volume of solution in liters}} = \frac{0.1 \, \text{mol}}{0.500 \, \text{L}} = 0.2 \, \text{M} $

Therefore, the molarity of the solution is 0.2 M. The correct option is D.

$\begin{aligned} & \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\ & \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \end{aligned}$

Let the weight of $\mathrm{CaCO}_3$ be $\mathrm{x}$ gm

$\therefore$ weight of $\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{~gm}$

Moles of $\mathrm{CaCO}_3$ decomposed $=$ moles of $\mathrm{CaO}$ formed

$\frac{\mathrm{x}}{100}=$ moles of $\mathrm{CaO}$ formed

$\therefore$ weight of $\mathrm{CaO}$ formed $=\frac{\mathrm{x}}{100} \times 56$

Moles of $\mathrm{MgCO}_3$ decomposed $=$ moles of $\mathrm{MgO}$ formed

$\frac{(2.21-x)}{84}=$ moles of $\mathrm{MgO}$ formed

$\therefore$ weight of $\mathrm{MgO}$ formed $=\frac{2.21-\mathrm{x}}{84} \times 40$

$\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152$

$\therefore \mathrm{x}=1.1886 \mathrm{~g}=$ weight of $\mathrm{CaCO}_3$

& weight of $\mathrm{MgCO}_3=1.0214 \mathrm{~g}$

Mole fraction of $\mathrm{C=\frac{n_C}{n_A+n_B+n_C}}$

The quantity that changes with temperature from the given options is Molarity.

Molarity, which is denoted with a capital M, measures the number of moles of solute dissolved in one liter of solution. It is expressed as:

$ Molarity (M) = \frac{moles \ of \ solute}{liters \ of \ solution} $

Since the volume of liquids expands or contracts with temperature, the volume of the solution will change as the temperature changes, which in turn will affect the molarity. For example, if the temperature of a solution increases, its volume may expand. If the number of moles of solute remains constant, the increased volume will result in a decreased molarity because the solute is now dispersed in a greater volume of solution.

In contrast, other measures of concentration such as:

Molality, which is measured as the number of moles of solute per kilogram of solvent (not affected by temperature since it's based on mass, not volume),

$ Molality (m) = \frac{moles \ of \ solute}{kilograms \ of \ solvent} $

Mole fraction, which is the ratio of the number of moles of one component to the total number of moles of all components in the solution (also not affected by temperature since the ratio of moles doesn't change with temperature),

$ Mole \ fraction \ (X) = \frac{moles \ of \ component}{total \ moles \ in \ mixture} $

and Mass percentage, which is the percentage by mass of a component in a mixture (again, not affected by temperature because it's based on mass),

$ Mass \ percentage \ $(%) $= \frac{mass \ of \ component}{total \ mass \ of \ mixture} \times 100\% $

do not change with temperature. Therefore, the correct answer to the question is:

Option B : Molarity

Statements given in (I) and (III) are correct, while statement given in (ii) is incorrect. The correct from is :

Manufacture of ammonia by Haber process is an exothermic reaction.

Determine Oxygen Mass:

The metal's mass is $ 10 \, \text{g} $. The resulting oxide weighs $ 11.6 \, \text{g} $. Thus, the mass of oxygen can be calculated as:

$ \text{Mass of oxygen} = 11.6 \, \text{g} - 10 \, \text{g} = 1.6 \, \text{g} $

Equivalent Mass of Oxygen:

It is given that the equivalent mass of oxygen is $ 8 \, \text{g} $.

Calculate Equivalent Mass of Metal (M):

Using the relationship for equivalent mass, which involves the mass of the substance and the mass of the reactive component (in this case, oxygen), we can set up the following equation:

$ \frac{\text{Equivalent mass of metal}}{\text{Equivalent mass of oxygen}} = \frac{\text{Mass of metal}}{\text{Mass of oxygen}} $

Plugging in the values:

$ \frac{x}{8} = \frac{10}{1.6} $

Solving for $ x $:

$ x = \frac{80}{1.6} = 50 $

Thus, the equivalent weight of the metal $ M $ is $ 50 $.

To find the remaining moles of $\mathrm{H}_2\mathrm{SO}_4$ in the flask, follow these steps:

Step 1: Calculate the Molar Mass

The molar mass of $\mathrm{H}_2\mathrm{SO}_4$ is determined by adding the atomic masses of its constituent elements:

Hydrogen ($H$): $2 \times 1 = 2$

Sulfur ($S$): $32$

Oxygen ($O$): $4 \times 16 = 64$

Thus, the total molar mass of $\mathrm{H}_2\mathrm{SO}_4$ is:

$ 2 + 32 + 64 = 98 \, \text{g/mol} $

Step 2: Convert Mass from mg to g

Given:

98 mg of $\mathrm{H}_2\mathrm{SO}_4$.

Since 1 g = 1000 mg, convert milligrams to grams:

$ 98 \, \text{mg} = \frac{98}{1000} \, \text{g} = 0.098 \, \text{g} $

Step 3: Determine Initial Molecules

The number of molecules in 98 mg of $\mathrm{H}_2\mathrm{SO}_4$ can be found using Avogadro’s number ($N_A = 6.02 \times 10^{23}$ molecules/mol):

98 grams of $\mathrm{H}_2\mathrm{SO}_4$ contains $6.02 \times 10^{23}$ molecules. Therefore, 0.098 grams will have:

$ \frac{6.02 \times 10^{23}}{1000} = 6.02 \times 10^{20} \text{ molecules} $

Step 4: Calculate Remaining Molecules

After removing $3.01 \times 10^{20}$ molecules, the remaining number of molecules is:

$ 6.02 \times 10^{20} - 3.01 \times 10^{20} = 3.01 \times 10^{20} \text{ molecules} $

Step 5: Convert Remaining Molecules to Moles

Using the proportion of molecules to moles:

$ 6.02 \times 10^{23} \text{ molecules} = 1 \text{ mol} $

Consequently, $3.01 \times 10^{20}$ molecules are:

$ \frac{3.01 \times 10^{20}}{6.02 \times 10^{23}} = 5 \times 10^{-4} \text{ moles} $

Thus, the remaining $\mathrm{H}_2\mathrm{SO}_4$ in the flask is $5 \times 10^{-4}$ moles.

The decomposition reaction is,

$ \underset{2 \text { moles }}{2 \mathrm{MMOO}_4} \longrightarrow \underset{1 \text { moles }}{\mathrm{K}_2 \mathrm{MnO}_4}+\underset{1 \text { moles }}{\mathrm{MnO}_2}+\underset{1 \text { moles }}{\mathrm{O}_2 \uparrow} $

For 0.1 moles of $\mathrm{KMnO}_4$

$ =0.1 / 2=0.05 \mathrm{moles} $

(same for $\mathrm{K}_2 \mathrm{MnO}_4$ and $\mathrm{MnO}_2$ )

Molar mass of products,

$ \begin{gathered} \mathrm{K}_2 \mathrm{MnO}_4=197 \mathrm{~g} / \mathrm{mol} \\ \mathrm{MnO}_2=87 \mathrm{~g} / \mathrm{mol} \end{gathered} $

Weight of $\mathrm{K}_2 \mathrm{MnO}_4=0.05 \times 197$

$ =9.85 \mathrm{~g} $

Weight of $\mathrm{MnO}_2=0.05 \times 87$

$ =4.35 \mathrm{~g} $

Weight of residue $=$ Weight of $\mathrm{K}_2 \mathrm{MnO}_4$ and $\mathrm{MnO}_2$

$ 9.85+4.35=14.2 \mathrm{~g} $

To determine the normality of the resulting solution with respect to $\mathrm{NO}_3^{-}$, we begin by examining the individual solutions and their normalities:

For $\mathrm{Ca(NO_3)_2}$:

Molarity ($M$): $\frac{M}{10}$

$n$-factor: 2 (because $\mathrm{Ca(NO_3)_2}$ provides two $\mathrm{NO}_3^{-}$ ions per formula unit)

Normality ($N$): Calculated as $N = M \times n\text{-factor} = \frac{1}{10} \times 2 = 0.2 \, \text{N}$

For $\mathrm{KNO_3}$:

Molarity ($M$): $\frac{M}{10}$

$n$-factor: 1 (because $\mathrm{KNO_3}$ provides one $\mathrm{NO}_3^{-}$ ion per formula unit)

Normality ($N$): Calculated as $N = M \times n\text{-factor} = \frac{1}{10} \times 1 = 0.1 \, \text{N}$

To find the normality of the mixed solution with respect to $\mathrm{NO}_3^{-}$, use the formula for mixing solutions:

$ N_{\text{result}} = \frac{V_1 N_1 + V_2 N_2}{V_1 + V_2} $

$V_1 = 100 \, \text{mL}$, $N_1 = 0.2 \, \text{N}$ for $\mathrm{Ca(NO_3)_2}$

$V_2 = 200 \, \text{mL}$, $N_2 = 0.1 \, \text{N}$ for $\mathrm{KNO_3}$

Plugging in the values:

$ N_{\text{result}} = \frac{100 \times 0.2 + 200 \times 0.1}{100 + 200} = \frac{20 + 20}{300} = \frac{40}{300} = 0.13 \, \text{N} $

Thus, the normality of the resulting solution with respect to $\mathrm{NO}_3^{-}$ is 0.13 N.

To determine the value of 'x', we need to analyze two reactions involving metal hydrogen carbonate ($ \text{MHCO}_3 $).

Reaction I: Decomposition of $ \text{MHCO}_3 $

$ 2 \, \text{MHCO}_3 \xrightarrow{\Delta} \text{CO}_2 + \text{M}_2\text{CO}_3 + \text{H}_2\text{O} $

In this reaction, 2 moles of $ \text{MHCO}_3 $ decompose to give 1 mole of $ \text{CO}_2 $.

For $ x \, \text{g} $ of $ \text{MHCO}_3 $, the number of moles is $ \frac{x}{84} $.

This results in $ \frac{x}{2 \times 84} $ moles of $ \text{CO}_2 $.

Reaction II: Reaction of $ \text{CO}_2 $ with $ \text{MOH} $

$ \text{CO}_2 + \text{MOH} \longrightarrow \text{MHCO}_3 $

In this reaction, 1 mole of $ \text{CO}_2 $ reacts with 1 mole of $ \text{MOH} $ to form 1 mole of $ \text{MHCO}_3 $.

From Reaction I, we have $ \frac{x}{2 \times 84} $ moles of $ \text{CO}_2 $.

According to the problem, this amount of $ \text{CO}_2 $ completely reacts with 0.2 moles of $ \text{MOH} $.

Set the moles of $ \text{CO}_2 $, $ \frac{x}{2 \times 84} $, equal to 0.2:

$ \frac{x}{2 \times 84} = 0.2 $

Solving for $ x $:

$ x = 0.2 \times 2 \times 84 = 33.6 $

Thus, the value of $ x $ is $ 33.6 $.

To determine the concentration of a 6% urea solution with a total weight of 1000 g, follow these steps:

Calculate the amount of urea:

The solution contains 6% urea by mass. Thus, the mass of urea is:

$ \text{Urea} = 1000 \, \text{g} \times \frac{6}{100} = 60 \, \text{g} $

The molar mass of urea (CH₄N₂O) is calculated using the atomic masses, which are:

Carbon (C): 12 u

Nitrogen (N): 14 u

Oxygen (O): 16 u

Hydrogen (H): 1 u

Therefore, the molar mass of urea is:

$ \text{Molar mass of urea} = 12 + (4 \times 1) + (2 \times 14) + 16 = 60 \, \text{g/mol} $

Convert the mass of urea to moles:

$ \text{Moles of urea} = \frac{60 \, \text{g}}{60 \, \text{g/mol}} = 1 \, \text{mol} $

Calculate the volume of water:

The remaining mass of the solution is water:

$ \text{Water} = 1000 \, \text{g} - 60 \, \text{g} = 940 \, \text{g} $

Using the density of water (1 g/mL), convert this mass to volume:

$ \text{Volume of water} = \frac{940 \, \text{g}}{1 \, \text{g/mL}} = 940 \, \text{mL} = 0.94 \, \text{L} $

Calculate the concentration of the urea solution:

The concentration in molarity ($ \text{mol/L} $) is given by:

$ \text{Concentration} = \frac{\text{Moles of urea}}{\text{Volume of solution in liters}} = \frac{1 \, \text{mol}}{0.94 \, \text{L}} = 1.0638 \, \text{M} $

Rounding to three significant figures gives:

$ \text{Concentration} \approx 1.064 \, \text{mol/L} $

Therefore, the concentration of the urea solution is approximately $1.064 \, \text{mol/L}$.

Given reaction is

$ \underset{\substack{\text { 1 mole }}}{\mathrm{Xe}(g)}+\underset{5 \text { moles }}{2 \mathrm{~F}_2(g)} \xrightarrow[7 \text { bar }]{873 \mathrm{~K}} \mathrm{XeF}_4(s) $

The ratio of $\mathrm{Xe}: \mathrm{Fe}_2$ is $1: 5$.

To find the concentration of nitric acid in mol/L, we'll use the provided data about its density and weight percentage.

Given:

Density of the solution = $1.5 \, \text{g/mL}$

Weight percentage of nitric acid = 68%

Calculate the mass of the solution in 1L:

Since the density is given as $1.5 \, \text{g/mL}$, the mass of 1 liter (1000 mL) of the solution would be:

$ \text{Mass of solution} = 1.5 \, \text{g/mL} \times 1000 \, \text{mL} = 1500 \, \text{g} $

Determine the mass of nitric acid in the solution:

The weight percentage indicates that 68% of the solution's mass is nitric acid. Therefore, the mass of nitric acid is:

$ \text{Mass of } \text{HNO}_3 = \frac{68}{100} \times 1500 \, \text{g} = 1020 \, \text{g} $

Calculate the molar mass of nitric acid ($\text{HNO}_3$):

Using the atomic masses provided:

$ \text{Molar mass of } \text{HNO}_3 = 1 \, (\text{H}) + 14 \, (\text{N}) + 16 \times 3 \, (\text{O}) = 63 \, \text{g/mol} $

Find the concentration in mol/L:

To find the number of moles of $\text{HNO}_3$ in 1L:

$ \text{Number of moles} = \frac{\text{Mass of } \text{HNO}_3}{\text{Molar mass } \text{HNO}_3} = \frac{1020 \, \text{g}}{63 \, \text{g/mol}} = 16.20 \, \text{mol/L} $

Therefore, the concentration of nitric acid is approximately $16.20 \, \text{mol/L}$.

Carbon $=92.3 / 12=7.7$

Hydrogen $=7.7 / 1=7.7$

Ratio $\mathrm{C}: \mathrm{H}=1$ : 1

So, emperical formula $=\mathrm{CH}$

In question, 39 g of hydrocarbon is given, it means 3 moies of CH is needed.

Also, $2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathbf{2 N a O H}+\mathbf{H}_2$

$2 \mathrm{H}_2 \mathrm{O}$ gives $1 \mathrm{~mol} \mathrm{H}_2$

so, $\frac{3}{2} \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ gives $\frac{1}{2} \times 1.5 \mathrm{~mol}$ of $\mathrm{H}_2$

$ =0.75 \mathrm{~mol} \mathrm{H}_2 $

So, all $\mathrm{H}_2 \mathrm{O}$ is consumed to liberate 0.75 mole of $\mathrm{H}_2$.

Thus, weight of $\mathrm{O}_2$ consumed

$ =\frac{15}{4} \times 32=120 \mathrm{~g} $

To determine the concentration of a $ \mathrm{CaCO}_3 $ solution in molarity (mol/L), given its concentration is 1000 ppm, we proceed as follows:

Mass of $\mathrm{CaCO}_3$ in 1 Liter:

Since 1000 ppm means 1000 mg of $\mathrm{CaCO}_3$ per liter, it converts to grams as:

$ \frac{1000 \times 1000 \, \text{mg}}{1000000} = 1 \, \text{g} $

Calculation of Moles:

The molar mass of $\mathrm{CaCO}_3$ is calculated using the atomic masses:

$\mathrm{Ca} = 40 \, \text{u}, \mathrm{C} = 12 \, \text{u}, \mathrm{O} = 16 \, \text{u}$.

Therefore, the molar mass of $\mathrm{CaCO}_3 = 40 + 12 + (16 \times 3) = 100 \, \text{g/mol}$.

Now, calculate the number of moles ($n$):

$ n = \frac{\text{Mass of } \mathrm{CaCO}_3}{\text{Molar mass}} = \frac{1}{100} = 0.01 \, \text{mol} $

Determine Molarity:

The molarity (concentration in mol/L) is given by:

$ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume of solution in L}} = \frac{0.01}{1} = 0.01 \, \text{mol/L} = 10^{-2} \, \text{mol/L} $

Thus, the concentration of the $\mathrm{CaCO}_3$ solution is $10^{-2} \, \text{mol/L}$.

The weight percent of chlorine in the compound is given as 55.0%. This implies that the weight percent of the metal is 45.0%.

Now, we need to find the molar mass of the compound. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L. We know that 100 mL (or 0.1 L) of the metal chloride vapor weighs 0.57 g. Therefore, the molar mass of the compound (MClx) is:

$M_{MCl_x} = \frac{0.57 \, \text{g}}{0.1 \, \text{L}} \times 22.4 \, \text{L/mol} = 127.68 \, \text{g/mol}$

Given the weight percent of the metal and chlorine, we can find the molar mass of the metal (MM) using the equation:

$MM = M_{MCl_x} \times \frac{45.0}{55.0} = 104.04 \, \text{g/mol}$

Subtracting the atomic mass of chlorine from the molar mass of the metal gives the molar mass of the metal:

$MM - x \cdot 35.5 = 104.04 \, \text{g/mol}$

Where 'x' is the number of chlorine atoms in the compound.

Solving this equation for 'x' gives:

$x = \frac{104.04 - MM}{35.5}$

We find that 'x' is approximately 2. Therefore, the molecular formula of the compound is MCl₂.

So, the correct answer is MCl₂.