# Question 44a79

Sep 3, 2017

WARNING! Long answer! The solution with the lowest vapour pressure
is d) 10 % $\text{NaCl}$.

#### Explanation:

The formula for vapour pressure lowering Δp is

color(blue)(barul|stackrel(" ")(Δp = χ_2p_1^@)|)

where

χ_2 = the mole fraction of the solute
${p}_{1}^{\circ} =$ the vapour pressure of the pure solvent

However, we must remember that vapour pressure lowering is a colligative property: it depends on the total number of particles in the solution.

Hence, when we are dealing with strong electrolytes, we must add the
van't Hoff $i$ factor to the equation.

The equation then becomes

color(blue)(barul|stackrel(" ")(Δp = iχ_2p_1^@)|)

Since the solvent is the same in every case, all we need to calculate is the value
of iχ_2.

Assume we have 1000 g of each solution.

a) 10 % ${\text{MgCl}}_{2}$

$\text{Mass of MgCl"_2 = "100 g}$

${\text{Moles of MgCl"_2 = 100 color(red)(cancel(color(black)("g MgCl"_2))) × "1 mol MgCl"_2/(95.21 color(red)(cancel(color(black)("g MgCl"_2)))) = "1.05 mol MgCl}}_{2}$

$\text{Mass of water = 900 g}$

$\text{Moles of water" = 900 color(red)(cancel(color(black)("g water"))) × "1 mol water"/(18.02 color(red)(cancel(color(black)("g water")))) = "49.9 mol water}$

χ_2 = n_2/(n_2 + n_1) = (1.05 color(red)(cancel(color(black)("mol"))))/((1.05 + 49.9) color(red)(cancel(color(black)("mol")))) = 1.05/51.0 = 0.0206

iχ_2 = 3 × 0.0206 = 0.0618

b) 10 % ${\text{CaCl}}_{2}$

$\text{Mass of CaCl"_2 = "100 g}$

${\text{Moles of CaCl"_2 = 100 color(red)(cancel(color(black)("g CaCl"_2))) × "1 mol CaCl"_2/(110.98 color(red)(cancel(color(black)("g CaCl"_2)))) = "0.901 mol CaCl}}_{2}$

χ_2 = n_2/(n_2 + n_1) = (0.901 color(red)(cancel(color(black)("mol"))))/((0.901 + 49.9) color(red)(cancel(color(black)("mol")))) = 0.901/50.8 = 0.0177

iχ_2 = 3 × 0.0177 = 0.0532

c) 10 % $\text{KCl}$

$\text{Mass of KCl = 100 g}$

$\text{Moles of KCl" = 100 color(red)(cancel(color(black)("g KCl"))) × "1 mol KCl"/(74.55 color(red)(cancel(color(black)("g KCl")))) = "1.34 mol KCl}$

χ_2 = n_2/(n_2 + n_1) = (1.34 color(red)(cancel(color(black)("mol"))))/((1.34 + 49.9) color(red)(cancel(color(black)("mol")))) = 0.901/51.3 = 0.0262

iχ_2 = 2 × 0.0262 = 0.0523

d) 10 % $\text{NaCl}$

$\text{Mass of NaCl = 100 g}$

$\text{Moles of NaCl" = 100 color(red)(cancel(color(black)("g NaCl"))) × "1 mol NaCl"/(58.44 color(red)(cancel(color(black)("g NaCl")))) = "1.71 mol NaCl}$

χ_2 = n_2/(n_2 + n_1) = (1.71 color(red)(cancel(color(black)("mol"))))/((1.71 + 49.9) color(red)(cancel(color(black)("mol")))) = 1.71/51.7 = 0.0262

iχ_2 = 2 × 0.0331 = 0.0663

The solution of 10 % $\text{NaCl}$ has the greatest value of iχ_2#, so it has the greatest vapor pressure lowering and the lowest vapour pressure.