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.

Step-by-step 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}}`

=
`(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}}`

=
`(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}}`

=
`(0.537 mol)/(0.537 mol + 13.9 mol)`

=
`(0.537 mol)/(14.416 mol)`

= 0.037

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

`(p^(o) - p^(solution))/(p^(o)) = i * x_(2)`

Here, i = 2 because KCl on dissociation produces 2 ions that is,
`K^(+)` and
`Cl^(-)`.

`(23.76 mm Hg - p^(solution))/(23.76 mm Hg) = 2 * 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.