Question

a solution contains 0.102 mol ca(no₃)₂ and 0.927 mol h₂o. calculate the vapour pressure of the solution at 55 °c. the vapour pressure of pure water at 55 °c is 118.1 torr. (assume that the solute completely dissociates).

Explanation:

Step1: Calculate the total number of moles of solute particles

Since \(Ca(NO_3)_2\) dissociates as \(Ca(NO_3)_2
ightarrow Ca^{2 +}+2NO_3^{-}\), the number of moles of solute particles \(n_{solute}=0.102\times3 = 0.306\) mol.

Step2: Calculate the mole fraction of water

The mole fraction of water \(x_{H_2O}=\frac{n_{H_2O}}{n_{H_2O}+n_{solute}}=\frac{0.927}{0.927 + 0.306}=\frac{0.927}{1.233}\approx0.752\)

Step3: Calculate the vapour pressure of the solution

Using Raoult's law \(P = x_{H_2O}\times P_{H_2O}^0\), where \(P_{H_2O}^0 = 118.1\) Torr.
\(P=0.752\times118.1\) Torr \(\approx88.8\) Torr

Answer:

\(88.8\) Torr