Chapter 14.2, Problem 30PP

Interpretation Introduction

Interpretation:

Amount of waterin 100 mL aqueoussolution of H₂SO₄ has to be calculated.

Concept introduction:

Mole fraction: Mole fraction defined as the number of moles of one component divided by total the number of moles in the mixture. Mole fraction is denoted as ‘ ${\chi }_i$ ’ where subscript ‘i’ represents the particular component. The sum of the mole fractions of the components in a mixture (or a solution) is unity. The mole fraction for the solvent ( ${\chi }_A$ )and the mole fraction for the solute ( ${\chi }_B$ )can be expressed as follows

Mole fraction ${\chi }_A=\frac{n_A}{ n_A+n_B}$ and ${\chi }_B=\frac{n_B}{ n_A+n_B}$ Where, ${\chi }_A$ and ${\chi }_B$ represent the mole fractions of each substance. n_A and n_Brepresent the number of moles of each substance.

$({\chi }_A+{\chi }_B)=1$

Answer to Problem 30PP

The mass of water is 27.54 g present in 100 mL aqueoussolution of H₂SO₄.

Explanation of Solution

Mole fraction ${\chi }_A=\frac{n_A}{ n_A+n_B}$ and ${\chi }_B=\frac{n_B}{ n_A+n_B}$ Where, ${\chi }_A$ and ${\chi }_B$ representthe mole fractions of H₂SO₄ and water respectively. n_A and n_Brepresent the number of moles of H₂SO₄ and water respectively.

(Data given: mole fractions of H₂SO₄ = 0.325)

$\mathrm{x}(H_{2}S{O}_4)=\frac{n(H_{2}S{O}_4)}{n(H_{2}S{O}_4)+n(H_{2}O)}$ ............................ (1)

Since, the volume of solution is 100 mL, since, the mass of solution is approximately 100g.

$\begin{array}{l}m\left(solution\right)=m\left(H_{2}S{O}_4\right)+m\left(H_{2}O\right)\hfill \\ m\left(H_{2}O\right)=n\left(H_{2}O\right)�M\left(H_{2}O\right)andm\left(H_{2}S{O}_4\right)=n\left(H_{2}S{O}_4\right)�M\left(H_{2}S{O}_4\right)\hfill \\ m\left(solution\right)=n\left(H_{2}S{O}_4\right)�M\left(H_{2}S{O}_4\right)+n\left(H_{2}O\right)�M\left(H_{2}O\right)\hfill \end{array}$

From the 1st equation: $n\left(H_{2}S{O}_4\right)=x\left(H_{2}S{O}_4\right)�\left(n\left(H_{2}S{O}_4\right)+n\left(H_{2}O\right)\right)$

Assume that n(H₂SO₄) is a, and n(H₂O) is b.

$\{m\left(solution\right)=a\times M(H_{2}S{O}_4)+b\times M(H_{2}O)\}$

$\mathrm{a}=x(H_{2}S{O}_4)\times \left(a+b\right)$ ........................(2)

From 2nd equation:

$a=\frac{0.325b}{1-0.325}$

If this to 1st equation, thus,

$\begin{array}{l}100=\frac{0.325b}{1-0.325}\times 98+b\times 18\\ 65.19\times b=100\\ b=n(H_{2}O)=1.53moles\end{array}$

The mass of water in this solution: $\mathrm{m}(H_{2}O)=n(H_{2}O)\times M(H_{2}O)$

$\begin{array}{l} =1.53moles\times 18g/mol\hfill \\ =27.54g\hfill \end{array}$