Some Basic Concepts of Chemistry

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]

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 $10^{-2} \, \text{mol/L}$.

To determine the percentage change in the molarity of the solution, we need to follow these steps:

1. Calculate the initial volume of the solution at $+90^{\circ} \mathrm{C}$.
2. Calculate the final volume of the solution at $+10^{\circ} \mathrm{C}$.
3. Determine the initial and final molarities of the solution.
4. Compute the percentage change in the molarity.

Given information:

• Mass of the solute (substance) = 50 g
• Mass of the solvent (water) = 1 kg = 1000 g
• Initial density = $1.1 \mathrm{~g} \mathrm{~cc}^{-1}$
• Final density = $1.15 \mathrm{~g} \mathrm{~cc}^{-1}$

Step 1: Calculate the initial volume of the solution at $+90^{\circ} \mathrm{C}$.

The total mass of the solution is:

$\text{Total mass} = 50 \, g + 1000 \, g = 1050 \, g$

Using the initial density, the initial volume is:

$V_{\text{initial}} = \frac{\text{Total mass}}{\text{Initial density}} = \frac{1050 \, g}{1.1 \, \mathrm{g/cc}} \approx 954.545 \, \mathrm{cc}$

Step 2: Calculate the final volume of the solution at $+10^{\circ} \mathrm{C}$.

Using the final density, the final volume is:

$V_{\text{final}} = \frac{\text{Total mass}}{\text{Final density}} = \frac{1050 \, g}{1.15 \, \mathrm{g/cc}} \approx 913.043 \, \mathrm{cc}$

Step 3: Determine the initial and final molarities of the solution.

Molarity (M) is defined as the number of moles of solute per liter of solution. Let's assume the molar mass (M_s) of the solute (substance) is required (denote it as M_s).

Number of moles of the solute:

$\text{Number of moles} = \frac{50 \, g}{M_s \, \mathrm{g/mol}} = \frac{50}{M_s} \, \mathrm{mol}$

Initial molarity:

$M_{\text{initial}} = \frac{\text{Number of moles}}{\text{Initial volume in liters}} = \frac{\frac{50}{M_s}}{\frac{954.545}{1000}} = \frac{50 \times 1000}{M_s \times 954.545} \approx \frac{50 \times 1.047}{M_s}$

Final molarity:

$M_{\text{final}} = \frac{\text{Number of moles}}{\text{Final volume in liters}} = \frac{\frac{50}{M_s}}{\frac{913.043}{1000}} = \frac{50 \times 1000}{M_s \times 913.043} \approx \frac{50 \times 1.095}{M_s}$

Step 4: Compute the percentage change in molarity.

Percentage change in molarity:

$\text{Percentage change} = \frac{M_{\text{final}} - M_{\text{initial}}}{M_{\text{initial}}} \times 100$

Substituting the values:

$\text{Percentage change} = \frac{\frac{50 \times 1.095}{M_s} - \frac{50 \times 1.047}{M_s}}{\frac{50 \times 1.047}{M_s}} \times 100 = \frac{50 \times (1.095 - 1.047)}{50 \times 1.047} \times 100$

$\text{Percentage change} = \frac{(1.095 - 1.047)}{1.047} \times 100 = \frac{0.048}{1.047} \times 100 \approx 4.58\%$

Since the closest provided option is 4.5, the correct answer is:

Option B: 4.5

Calgon is a water softener. The molecular formula of calgon is $\mathrm{Na}_6 \mathrm{P}_6 \mathrm{O}_{18}$. Thus, the empirical formula of calgon is $\mathrm{NaPO}_3$.

Law of definite proportion states that the individual elements that constitute a chemical compound are always present in a fixed mass ratio.

(a) The ratio of oxygen in $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{O}_2$ with respect to fixed mass of hydrogen atom is a whole number. This is the statement of "law of multiple proportion."

(b) As the ratio of elements is fixed, so percentage of oxygen in $\mathrm{H}_2 \mathrm{O}$ is constant or fixed irrespective of the source. This statement is with reference to law of definite proportion. Thus, this is the correct option.

(c) Avogadro's law states that the "equal volume of all gases at the same temperature and pressure should contain equal number of molecules."

(d) "Mass can neither be created nor be destroyed" is the statement of "law of conservation of mass."

The balanced equation is

$2 \mathrm{Pb}_3 \mathrm{O}_4 \longrightarrow 6 \mathrm{PbO}+\mathrm{O}_2$

So, value of $x$ and $y$ in the given reactions are 2 and 6 respectively.

$\begin{aligned} M & =15.9 \mathrm{~g} \text { (Given }) \\ V & =10 \mathrm{~mL} \end{aligned}$

$\begin{aligned} & \text { Density }=\frac{\text { Mass }}{\text { Volume }}=\frac{15.9 \mathrm{~g}}{10 \mathrm{~mL}} \\ & \text { Density }=1.59 \mathrm{~g} \mathrm{~mL}^{-1} . \end{aligned}$

Volume of solution

$\begin{aligned} &=250 \mathrm{~cm}^3=250 \mathrm{~mL} \\ & =250 \times 10^{-3} \mathrm{~L} \end{aligned}$

Mass of oxalic acid $=0.63 \mathrm{~g}$

Molar mass of oxalic acid $=126$

Moles $=0.63 / 126=5 \times 10^{-3}$ moles

$\begin{aligned} \text { Molarity } & =\frac{\text { Number of moles of solution }}{\text { Volume of solution in litre }} \\ M & =\frac{5 \times 10^{-3}}{250 \times 10^{-3}}=0.02 \mathrm{M}\end{aligned}$

Normality = Basicity $\times$ Molarity

Basicity of oxalic acid = 2

$N=2 \times 0.02=0.04 \mathrm{~N}$.

1 mole of benzene produces 1 mole of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$ $78 \mathrm{~g}$ of benzene produces $120 \mathrm{~g}$ of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$

$\therefore 7.8 \mathrm{~g}$ of benzene produce

$\begin{aligned} & =\frac{120}{78} \times 7.8=12 \mathrm{~g} \\ \text { Percentage yield } & =\frac{\text { Actual yield }}{\text { Theoretical yield }} \times 100 \\ & =\frac{8.4}{12} \times 100=70 \% \end{aligned}$

$\mathrm{\mathop {2{C_6}{H_6}}\limits_{Benzene} + 15{O_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 12C{O_2} + 6{H_2}O}$

2 moles of benzene produce 12 moles of $\mathrm{CO}_2$.

$\therefore 1 \mathrm{~mol}$ of benzene produce 6 moles of $\mathrm{CO}_2$

$\begin{aligned} \text { Mol } & =\frac{\text { Given mass }}{\text { Molecular mass }} \\ 6 & =x / 44 \Rightarrow x=264 \mathrm{~g} \end{aligned}$

$1 \mathrm{~mol}$ of benzene produces $264 \mathrm{~g}$ of $\mathrm{CO}_2$.

Mass of sample of alloy = 12 g

$\%$ of metal $X$ in sample $=20$

$\therefore$ If $x$ is the mass of metal $X$ in sample

$\begin{aligned} \frac{x}{12} \times 100 & =20 \\ x & =2.4 \mathrm{~g} \end{aligned}$

If $y$ is the mass of metal $Y$ in sample, then

$y=12-2.4=9.6 \mathrm{~g}$

Number of atoms of $X=\frac{6.022 \times 10^{23} \times 2.4}{40}$

$=3.61 \times 10^{22}$

Ratio of atoms of $X$ and $Y=2: 5$

Number of atoms of $Y=\frac{3.61 \times 10^{22} \times 5}{2}$

$=9.025 \times 10^{22}$ atoms.

$9.025 \times 10^{22}$ atoms of $Y$ are present in $9.6 \mathrm{~g}$.

$6.022 \times 10^{23}$ atoms of $Y$ are present in $\frac{9.5}{9.025 \times 10^{22}} \times 6.022 \times 10^{23}=64 \mathrm{~g}$

Atomic mass of $Y=64 \mathrm{~amu}$.

Moles of sulphuric acid, $n_A=0.2$

Volume of water, $V=250$ mL

Concentration of the solution/molarity

$=\frac{n_A}{V(L)}\times1000=\frac{0.2}{250}\times1000=0.8$ M

Molarity (molar concentration) : the number of moles of solute (sulphuric acid) per litre of solution (water).

Let $\mathrm{CaCO}_3(s)$ be consumed completely.

$\begin{aligned} & \mathrm{CaCO}_3(s)+2 \mathrm{HCl}(a q .) \longrightarrow \mathrm{CaCl}_2(a q) \\ & \begin{array}{lll} \text { Carcium } & \text { Hydrochloric } & \text { Calcium } \\ \text { chloride } \end{array} \\ & +\mathrm{H}_2 \mathrm{O}(l)+\mathrm{CO}_2(g) \\ & \therefore 100 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { gives }=44 \mathrm{~g} \text { of } \mathrm{CO}_2 \text { gas } \\ & \therefore 20 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { will give }=\frac{44}{100} \times 20 \mathrm{~g} \text { of } \mathrm{CO}_2 \\ & =8.8 \mathrm{~g} \text { of } \mathrm{CO}_2 \\ \end{aligned}$

Now, $\mathrm{HCl}$, be completely consumed.

$73 \mathrm{~g}$ of $\mathrm{HCl}$ give $=44 \mathrm{~g}$ of $\mathrm{CO}_2$ gas

$\therefore 20 \mathrm{~g}$ of $\mathrm{HCl}$ gives $=\frac{44}{73} \times 20 \mathrm{~g}$ of $\mathrm{CO}_2=12.05 \mathrm{~g}$ of $\mathrm{CO}_2$

Hence, $\mathrm{CaCO}_3$ give least amount of $\mathrm{CO}_2$ gas. So, $\mathrm{CaCO}_3$ is limiting reactant

$\therefore$ Amount of $\mathrm{CO}_2$ formed will be $8.8 \mathrm{~g}$.