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.

We have the following masses:

• A: 1.8 mg H₂O

• B: 9.8 mg H₂SO₄

• C: 1.8 mg C

• D: 5.85 mg NaCl

Since $ 1 \, \text{mg} = 10^{-3} \, \text{g} $:

$ \begin{aligned} \text{A: } & 1.8 \, \text{mg} = 1.8 \times 10^{-3} \, \text{g} \\ \text{B: } & 9.8 \, \text{mg} = 9.8 \times 10^{-3} \, \text{g} \\ \text{C: } & 1.8 \, \text{mg} = 1.8 \times 10^{-3} \, \text{g} \\ \text{D: } & 5.85 \, \text{mg} = 5.85 \times 10^{-3} \, \text{g} \end{aligned} $

Number of atoms $ = (\text{number of moles of molecules}) \times (\text{atoms per molecule}) \times N_A $

i.e.

$ \text{Number of atoms} = \frac{\text{Given mass}}{\text{Molar mass}} \times (\text{atoms per molecule}) \times N_A $

(A) Water (H₂O)

Molar mass of H₂O = 18 g/mol

Atoms per molecule = 3 (2 H + 1 O)

$ \text{Moles of H₂O} = \frac{1.8 \times 10^{-3}}{18} = 1 \times 10^{-4} $

$ \text{Number of atoms} = 1 \times 10^{-4} \times 3 N_A = 3 \times 10^{-4} N_A $

So, A → III

(B) Sulphuric acid (H₂SO₄)

Molar mass of H₂SO₄ = 98 g/mol

Atoms per molecule = 7 (2 H + 1 S + 4 O)

$ \text{Moles of H₂SO₄} = \frac{9.8 \times 10^{-3}}{98} = 1 \times 10^{-4} $

$ \text{Number of atoms} = 1 \times 10^{-4} \times 7 N_A = 7 \times 10^{-4} N_A $

So, B → IV

(C) Carbon (C)

Molar mass of carbon = 12 g/mol

Each atom = 1 atom per molecule

$ \text{Moles of C} = \frac{1.8 \times 10^{-3}}{12} = 1.5 \times 10^{-4} $

$ \text{Number of atoms} = 1.5 \times 10^{-4} N_A $

So, C → II

(D) Sodium chloride (NaCl)

Molar mass = 58.5 g/mol

Atoms per formula unit = 2 (Na + Cl)

$ \text{Moles of NaCl} = \frac{5.85 \times 10^{-3}}{58.5} = 1 \times 10^{-4} $

$ \text{Number of atoms} = 1 \times 10^{-4} \times 2 N_A = 2 \times 10^{-4} N_A $

So, D → I

$ \boxed{ A - III, \quad B - IV, \quad C - II, \quad D - I } $

Option C

$ \text{Total number of atoms} = (\text{Number of moles}) \times N_A \times (\text{Atomicity}) $

Let's calculate this for each option.

A. 2 g of $\mathrm{O}_2$ gas

1. Molar mass of $\mathrm{O}_2$: The atomic mass of O is 16 u. So, the molar mass of $\mathrm{O}_2$ is $2 \times 16 = 32 \ \text{g mol}^{-1}$.

2. Number of moles ($n$):

   $ n = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{2 \ \text{g}}{32 \ \text{g mol}^{-1}} = \frac{1}{16} \ \text{mol} $

3. Atomicity of $\mathrm{O}_2$: One molecule of $\mathrm{O}_2$ contains 2 oxygen atoms. So, the atomicity is 2.

4. Total number of atoms:

   $ \text{Atoms} = n \times N_A \times \text{Atomicity} = \frac{1}{16} \times N_A \times 2 = \frac{2}{16} N_A = \frac{1}{8} N_A $

B. 4 g of $\mathrm{SO}_2$ gas

1. Molar mass of $\mathrm{SO}_2$: Atomic mass of S is 32 u and O is 16 u. So, the molar mass of $\mathrm{SO}_2$ is $32 + (2 \times 16) = 32 + 32 = 64 \ \text{g mol}^{-1}$.

2. Number of moles ($n$):

   $ n = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \ \text{g}}{64 \ \text{g mol}^{-1}} = \frac{1}{16} \ \text{mol} $

3. Atomicity of $\mathrm{SO}_2$: One molecule of $\mathrm{SO}_2$ contains 1 sulfur atom and 2 oxygen atoms, making a total of $1+2=3$ atoms. The atomicity is 3.

4. Total number of atoms:

   $ \text{Atoms} = n \times N_A \times \text{Atomicity} = \frac{1}{16} \times N_A \times 3 = \frac{3}{16} N_A $

C. 1400 mL of $\mathrm{O}_2$ at STP

1. Molar volume at STP: We know that 1 mole of any gas at STP occupies 22400 mL (or 22.4 L).

2. Number of moles ($n$):

   $ n = \frac{\text{Given volume}}{\text{Molar volume at STP}} = \frac{1400 \ \text{mL}}{22400 \ \text{mL mol}^{-1}} = \frac{14}{224} = \frac{1}{16} \ \text{mol} $

3. Atomicity of $\mathrm{O}_2$: As we found earlier, the atomicity is 2.

4. Total number of atoms:

   $ \text{Atoms} = n \times N_A \times \text{Atomicity} = \frac{1}{16} \times N_A \times 2 = \frac{2}{16} N_A = \frac{1}{8} N_A $

D. 0.05 L of He at STP

1. Molar volume at STP: 1 mole of any gas at STP occupies 22.4 L.

2. Number of moles ($n$):

   $ n = \frac{\text{Given volume}}{\text{Molar volume at STP}} = \frac{0.05 \ \text{L}}{22.4 \ \text{L mol}^{-1}} \approx 0.0022 \ \text{mol} $

3. Atomicity of He: Helium (He) is a noble gas and exists as single atoms. So, its atomicity is 1.

4. Total number of atoms:

   $ \text{Atoms} = n \times N_A \times \text{Atomicity} = \frac{0.05}{22.4} \times N_A \times 1 = \frac{0.05}{22.4} N_A $

E. 0.0625 mol of $\mathrm{H}_2$ gas

1. Number of moles ($n$): This is given directly as $0.0625 \ \text{mol}$. Let's convert this decimal to a fraction to make comparison easier: $0.0625 = \frac{625}{10000} = \frac{1}{16}$. So, $n = \frac{1}{16} \ \text{mol}$.

2. Atomicity of $\mathrm{H}_2$: One molecule of $\mathrm{H}_2$ contains 2 hydrogen atoms. The atomicity is 2.

3. Total number of atoms:

   $ \text{Atoms} = n \times N_A \times \text{Atomicity} = \frac{1}{16} \times N_A \times 2 = \frac{2}{16} N_A = \frac{1}{8} N_A $

Conclusion

Let's summarise the total number of atoms we calculated for each option:

• A: $\frac{1}{8} N_A$ atoms

• B: $\frac{3}{16} N_A$ atoms

• C: $\frac{1}{8} N_A$ atoms

• D: $\frac{0.05}{22.4} N_A$ atoms

• E: $\frac{1}{8} N_A$ atoms

By comparing these results, we can clearly see that options A, C, and E contain the same number of atoms.

Therefore, the correct choice is the one that lists A, C, and E.

The correct answer from the options given below is Option D.

Let the oxide of iron contain 100 g of the compound.

Then,

Mass of Fe = 69.9 g

Mass of O = 100 − 69.9 = 30.1 g

Calculate moles of each element

$ \text{Moles of Fe} = \frac{69.9}{56} = 1.248 $

$ \text{Moles of O} = \frac{30.1}{16} = 1.881 $

Find the simplest whole-number ratio

Divide each by the smaller number (1.248):

$ \text{Ratio of Fe:O} = \frac{1.248}{1.248} : \frac{1.881}{1.248} = 1 : 1.51 $

This ratio is approximately 1 : 1.5.

To remove the fraction, multiply both by 2 : $ \text{Fe}_2\text{O}_3 $

In $\mathrm{Fe_2O_3}$,

Mass of Fe = $2 \times 56 = 112$ g

Mass of O = $3 \times 16 = 48$ g

Total mass = 160 g

Percentage of Fe = $\dfrac{112}{160} \times 100 = 70\%$ (approximately 69.9%)

Therefore, the empirical formula of the oxide is

$ \boxed{\mathrm{Fe_2O_3}} $

Correct Option: B.

The reaction is:

$ \mathrm{Zn} + \mathrm{H_2SO_4} \rightarrow \mathrm{ZnSO_4} + \mathrm{H_2} $

From the balanced equation,

$ 1 \text{ mol Zn } \text{ reacts with } 1 \text{ mol } \mathrm{H_2SO_4} \text{ and gives } 1 \text{ mol } \mathrm{H_2} $

Now let us find the number of moles of each reactant.

1. Amount of $\mathrm{H_2SO_4}$ present

Volume of acid solution $= 50 \, \mathrm{mL}$

Density $= 1.3 \, \mathrm{g\,mL^{-1}}$

So, mass of acid solution is

$ \text{Mass} = \text{Volume} \times \text{Density} = 50 \times 1.3 = 65 \, \mathrm{g} $

Purity $= 50\%$

Hence, mass of pure $\mathrm{H_2SO_4}$ is

$ \frac{50}{100} \times 65 = 32.5 \, \mathrm{g} $

Molar mass of $\mathrm{H_2SO_4}$:

$ 2(1) + 32 + 4(16) = 98 \, \mathrm{g\,mol^{-1}} $

So, moles of $\mathrm{H_2SO_4}$ are

$ \frac{32.5}{98} = 0.3316 \text{ mol} $

2. Amount of zinc

Mass of Zn $= 20 \, \mathrm{g}$

Molar mass of Zn $= 65 \, \mathrm{g\,mol^{-1}}$

So, moles of Zn are

$ \frac{20}{65} = 0.3077 \text{ mol} $

3. Limiting reagent

Since the reaction ratio is $1:1$:

• $\mathrm{H_2SO_4}$ available $= 0.3316$ mol

• Zn available $= 0.3077$ mol

Zn is smaller in amount, so zinc is the limiting reagent.

Therefore, moles of $\mathrm{H_2}$ formed are

$ 0.3077 \text{ mol} $

4. Volume of hydrogen at STP

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

So, volume of $\mathrm{H_2}$ is

$ 0.3077 \times 22.4 = 6.892 \, \mathrm{L} $

$ \boxed{6.892 \, \mathrm{L}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

$ \mathrm{Ag_2CO_3}(s) \xrightarrow{\Delta} 2\mathrm{Ag}(s) + \mathrm{CO_2}(g) + \frac{1}{2}\mathrm{O_2}(g) $

From the chemical equation, we can see that carbon dioxide ($ \mathrm{CO_2} $) and oxygen ($ \mathrm{O_2} $) are gases and will easily escape. The only solid residue left behind is silver ($ \mathrm{Ag} $).

Next, let us calculate the molar mass of silver carbonate ($ \mathrm{Ag_2CO_3} $) by adding the molar masses of its constituent elements:

$ \text{Molar mass of } \mathrm{Ag_2CO_3} = (2 \times 108) + 12 + (3 \times 16) $

$ \text{Molar mass of } \mathrm{Ag_2CO_3} = 216 + 12 + 48 = 276 \mathrm{~g} \mathrm{~mol}^{-1} $

Now, we can find the number of moles of silver carbonate provided in the question. The given mass is $ 2.76 \mathrm{~g} $.

$ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} $

$ \text{Moles of } \mathrm{Ag_2CO_3} = \frac{2.76 \mathrm{~g}}{276 \mathrm{~g} \mathrm{~mol}^{-1}} = 0.01 \mathrm{~mol} $

Using the stoichiometry from our balanced chemical equation, we know that heating $ 1 \mathrm{~mol} $ of $ \mathrm{Ag_2CO_3} $ yields $ 2 \mathrm{~mol} $ of solid $ \mathrm{Ag} $ residue.

Therefore, if we decomposition $ 0.01 \mathrm{~mol} $ of $ \mathrm{Ag_2CO_3} $, the number of moles of silver residue produced will be:

$ \text{Moles of residue } (\mathrm{Ag}) = 2 \times 0.01 \mathrm{~mol} = 0.02 \mathrm{~mol} $

Finally, we convert these moles of silver back to mass in grams. We know the given molar mass of $ \mathrm{Ag} $ is $ 108 \mathrm{~g} \mathrm{~mol}^{-1} $.

$ \text{Mass of } \mathrm{Ag} = \text{Moles of } \mathrm{Ag} \times \text{Molar mass of } \mathrm{Ag} $

$ \text{Mass of } \mathrm{Ag} = 0.02 \mathrm{~mol} \times 108 \mathrm{~g} \mathrm{~mol}^{-1} = 2.16 \mathrm{~g} $

The mass of the solid residue obtained is $ 2.16 \mathrm{~g} $.

Correct Option: Option B

For a $10\%$ by weight $(w/w)$ aqueous urea solution, let us take $100 \, \text{g}$ of solution.

Then:

• Mass of urea $= 10 \, \text{g}$

• Mass of water $= 90 \, \text{g}$

Now find the molar masses.

Urea has formula $NH_2CONH_2$ or $CO(NH_2)_2$

$ M(\text{urea}) = 12 + 16 + 2(14) + 4(1) = 60 \, \text{g mol}^{-1} $

Moles of urea:

$ n_{\text{urea}} = \frac{10}{60} = \frac{1}{6} \approx 0.167 $

Molar mass of water:

$ M(\text{water}) = 2(1) + 16 = 18 \, \text{g mol}^{-1} $

Moles of water:

$ n_{\text{water}} = \frac{90}{18} = 5 $

Now, mole fraction of water is

$ X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{urea}}} $

$ X_{\text{water}} = \frac{5}{5 + \frac{1}{6}} = \frac{5}{\frac{31}{6}} = \frac{30}{31} $

$ X_{\text{water}} \approx 0.967 $

So, the correct answer is:

$ \boxed{0.967} $

Option D

Let us find the total number of atoms in each case.

We only need to compare, so first calculate the number of moles of atoms in each sample.

(A) 2 moles of cyclohexane

Cyclohexane has formula:

$ \mathrm{C_6H_{12}} $

So, one molecule of cyclohexane contains:

$ 6 + 12 = 18 \text{ atoms} $

Therefore, in $2$ moles of cyclohexane, moles of atoms are:

$ 2 \times 18 = 36 \text{ moles of atoms} $

(B) 684 g of sucrose

Sucrose has formula:

$ \mathrm{C_{12}H_{22}O_{11}} $

Total atoms in one molecule:

$ 12 + 22 + 11 = 45 $

Molar mass of sucrose:

$ 12 \times 12 + 22 \times 1 + 11 \times 16 $

$ = 144 + 22 + 176 = 342 \text{ g mol}^{-1} $

Given mass $= 684$ g, so moles of sucrose:

$ \frac{684}{342} = 2 \text{ mol} $

Hence, moles of atoms:

$ 2 \times 45 = 90 \text{ moles of atoms} $

(C) 90.8 L of dihydrogen at STP

At STP, $1$ mole of any gas occupies $22.4$ L.

So moles of $\mathrm{H_2}$:

$ \frac{90.8}{22.4} \approx 4.05 \text{ mol} $

Each molecule of $\mathrm{H_2}$ has $2$ atoms.

Therefore, moles of atoms:

$ 4.05 \times 2 = 8.10 \text{ moles of atoms} $

Comparison

• (A) $= 36$ moles of atoms

• (B) $= 90$ moles of atoms

• (C) $= 8.10$ moles of atoms

So the correct order is:

$ \mathrm{B} > \mathrm{A} > \mathrm{C} $

Therefore, the correct option is:

$ \boxed{\text{Option D}} $

At STP, according to Avogadro’s law, $1$ mole of any gas occupies $22.4 \, \text{L}$.

So, number of moles of $\mathrm{SO_2}$ is

$ n=\frac{V}{22.4}=\frac{1.4187}{22.4} $

$ n=0.0633 \text{ mol} $

Now, number of molecules

$ = n \times N_A $

where $N_A = 6.022 \times 10^{23}$

So,

$ \text{Number of molecules} = 0.0633 \times 6.022 \times 10^{23} $

$ = 3.812 \times 10^{22} $

Hence, the correct answer is:

$ \boxed{0.0633 \text{ mol ; } 3.812 \times 10^{22} \text{ molecules}} $

So, Option B is correct.

Let the hydrocarbon be $C_2H_x$ since it has two carbon atoms.

Given, the mass percentage ratio of $C:H$ is $12:1$.

So, in the hydrocarbon,

• mass of carbon $= 2 \times 12 = 24$

• mass of hydrogen $= x \times 1 = x$

Now,

$ 24 : x = 12 : 1 $

So,

$ \frac{24}{x} = \frac{12}{1} $

$ 24 = 12x $

$ x = 2 $

Hence, the hydrocarbon is

$ C_2H_2 $

Its molar mass is

$ (2 \times 12) + (2 \times 1) = 24 + 2 = 26 \text{ g mol}^{-1} $

Number of moles of $C_2H_2$ in $3.38$ g:

$ \text{moles of } C_2H_2 = \frac{3.38}{26} = 0.13 \text{ mol} $

Complete combustion of acetylene:

$ 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O $

From the equation,

$ 1 \text{ mol } C_2H_2 \rightarrow 2 \text{ mol } CO_2 $

Therefore,

$ 0.13 \text{ mol } C_2H_2 \rightarrow 0.26 \text{ mol } CO_2 $

Molar mass of $CO_2$:

$ 12 + (2 \times 16) = 44 \text{ g mol}^{-1} $

Mass of $CO_2$ formed:

$ 0.26 \times 44 = 11.44 \text{ g} $

So, the correct answer is

$ \boxed{11.44 \text{ g}} $

Option B.

The reaction of iron with steam is:

$ 3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 $

We are given:

$ \text{Mass of steam} = 18 \text{ g} $

Molar mass of water:

$ H_2O = 2(1) + 16 = 18 \text{ g mol}^{-1} $

So, moles of steam are:

$ \text{Moles of } H_2O = \frac{18}{18} = 1 \text{ mol} $

From the balanced equation:

$ 4 \text{ mol } H_2O \text{ react with } 3 \text{ mol } Fe $

Therefore, for $1$ mol $H_2O$:

$ \text{Moles of Fe required} = \frac{3}{4} $

Now, molar mass of iron is:

$ 56 \text{ g mol}^{-1} $

So, mass of iron reacted:

$ \text{Mass of Fe} = \frac{3}{4} \times 56 = 42 \text{ g} $

Hence, the mass of iron converted into $Fe_3O_4$ is:

$ \boxed{42 \text{ g}} $

So, the correct option is:

$ \boxed{\text{Option D}} $

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 $2.79 \, \text{m}.$

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