Question

A KCl solution is prepared by dissolving 40.0 g KCl (molar mass = 74.55 g/mol) in 250.0 g of water (molar mass = 18.01 g/mol) at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?

a) 20.5 mm Hg
b) 22.1 mm Hg
c) 22.9 mm Hg
d) 24.7 mm Hg

Answer 1

Answer: Option (B) is the correct answer.

Explanation:

The given data is as follows.

    mass of KCl = 40 g,           mass of water = 250.0 g

Hence, number of moles of KCl will be calculated as follows.

             No. of moles = $\frac{mass}{\text{molar mass}}$

                                  = $\frac{40 g}{74.55 g/mol}$

                                  = 0.537 mol

Number of moles of water will be calculated as follows.

            No. of moles = $\frac{mass}{\text{molar mass}}$  

                                 = $\frac{250 g}{18.01 g/mol}$

                                 = 13.9 mol

Also, mole fraction of KCl will be calculated as follows.

                 $x_{KCl} = \frac{\text{moles of KCl}}{\text{total no. of moles}}$

                              = $\frac{0.537 mol}{0.537 mol + 13.9 mol}$

                              = $\frac{0.537 mol}{14.416 mol}$

                              = 0.037

Hence, calculate the vapor pressure of the solution as follows.

             $\frac{p^{o} - p^{solution}}{p^{o}} = i \times x_{2}$

Here, i = 2 because KCl on dissociation produces 2 ions that is, $K^{+}$ and $Cl^{-}$.

            $\frac{23.76 mm Hg - p^{solution}}{23.76 mm Hg} = 2 \times 0.037$

                  $p^{solution}$ = 22.1 mm Hg

Thus, we can conclude that the vapor pressure of the given solution is 22.1 mm Hg.