Chapter 16, Problem 79A

Interpretation Introduction

Interpretation: The mole fraction of $0.150m$

  $\text{NaCl}$ in an aqueous solution is to be determined.

Concept Introduction:

The expression for the mole fraction is as follows:

  $X_a=\frac{n_a}{n_a+n_b}$

Here, $n_a$ are the moles of solvent, $n_b$ are the moles of solute, and $X_a$ is the mole fraction of solute.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

The expression for molality is as follows:

  $\text{Molality}=\frac{\text{mole}\text{of}\text{solute}}{\text{mass}\text{of}\text{solvent }\left(\text{kg}\right)}$   ...... (2)

Answer to Problem 79A

The mole fraction of $0.150m$

  $\text{NaCl}$ solute is $0.0027$ and the solvent is $0.9973$ .

Explanation of Solution

Given:

Molality of $\text{NaCl}$ is $0.150m$ .

Solute is $\text{NaCl}$ and the solvent is water.

The moles of solvent should be calculated,

Weight of the solvent is $1000\text{g}$

Molar mass of water is $18\text{g/mol}$

Substitute, $1000\text{g}$ for the mass of water and $18\text{g/mol}$ for molar mass of solvent in equation (1).

  $\begin{array}{c}\text{moles}=\frac{\text{ 1000 g}}{18\text{ g/mol}}\\ =55.5\text{ mol}\end{array}$

Substitute, $55.5\text{mol}$ for moles of $\text{NaCl}$ and $0.150m$ for molality in equation (1).

  $\begin{array}{c}\text{Molality}=\frac{\text{mole}\text{of}\text{solute}}{\text{mass}\text{of}\text{solvent }\left(\text{kg}\right)}\\ \text{moles of solute }\left({\text{n}}_a\right)=\text{0}\text{.150}\times \text{55}\text{.5}\\ \text{= 8}\text{.32 mol}\end{array}$

Substitute $\text{8}\text{.32 mol}$ for $n_a$ and $55.5$ for $n_b$ in the formula of mole fraction to calculate the mole fraction of solute, $X_a$ .

  $\begin{array}{c}{X}_a=\frac{8.32\text{mol}}{8.32\text{mol}+55.5\text{mol}}\\ =0.13\end{array}$

The sum of the mole fraction of solute and solvent is 1. Hence, the mole fraction of solvent is,

  $1-0.13=0.87$

Conclusion

Therefore, the mole fraction of $0.150m$

  $\text{NaCl}$ solute is $0.13$ and the solvent is $0.87$ .