A solution contains 0.115 mol

$$H_2O$$

and an unknown number of moles of sodium ch lo ride. The vapor pressure of the solution at

$$30 ^ { \circ } \mathrm { C }$$

is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

The number of moles of water is $n_{H_2O} = 0.115 \mathrm{mol}$

The vapor pressure of the sodium chloride solution at $30^{\circ} \mathrm{C}$ is $P_{solution} = 25.7 \mathrm{torr}$

The vapor pressure of pure water is $P^{o}_{H_2O} = 31.8 \mathrm{torr}$

Let us calculate the mass of sodium chloride in the solution.

First, we have to find the mole fraction of water.

$$\begin{align*} P_{solution} &= X_{H_2O} \cdot P^{o}_{H_2O} \\ X_{H_2O} &= \frac { P_{solution} } { P^{o}_{H_2O} }\\ &= \frac { 25.7 \mathrm{torr} } { 31.8 \mathrm{torr} }\\ &= 0.808 \end{align*}$$

Now, we can calculate the number of moles of $\mathrm{NaCl}$.

• Since $\mathrm{NaCl}$ is strong electrolyte, it is completely ionized to form 2 $\mathrm{mol}$ of dissolved particles. Therefore

$$\begin{align*} X_{H_2O} &= \frac { n_{H_2O} } { n_{H_2O} + 2 \cdot n_{NaCl} }\\ 0.808 &= \frac {0.115 \mathrm{mol} } { 0.115 \mathrm{mol} + 2 \cdot n_{NaCl} }\\ 0.808 \cdot (0.115 \mathrm{mol} + 2 \cdot n_{NaCl}) &= 0.115 \mathrm{mol} \\ 0.093 \mathrm{mol} + 1.616 \cdot n_{NaCl} &= 0.115 \mathrm{mol} \\ 1.616 \cdot n_{NaCl} &= 0.022 \mathrm{mol}\\ n_{NaCl} &= 0.0136 \mathrm{mol} \end{align*}$$

The vapor pressure of the mixture and pure water is given. Use the Raoult's Law to find the mole fraction of water in the solution (with $\text{X}_{\text{H}_2\text{O}}$ more ration, mixture partial pressure $P_{vap}$ and pure solution partial pressure $P_{\text{H}_2\text{O}}$):

$$X_{\text{H}_2\text{O}}=\frac{P_{vap}}{P_{\text{H}_2\text{O}}}$$

$$\therefore X_{\text{H}_2\text{O}}=\frac{25.7}{31.8}$$

$$\therefore X_{\text{H}_2\text{O}}=0.8081761$$