Chapter 14, Problem 114A

Interpretation Introduction

Interpretation:

The solubility of argon at pressure 15 atm have to be calculated.

Concept introduction:

Solubility: Solubility can be expressed as maximum amount of solute that can dissolve in a known amount of solvent at a given temperature and pressure.

Henry’s Law: The solubility of a gas in liquid varies proportional to the pressure of the gas which is above the liquid at a given temperature.

Answer to Problem 114A

The solubility of argon at pressure 15 atmis 82.5 mg/100 gm water.

Explanation of Solution

The expression of Henry’s Law is $\frac{S_1}{P_1}=\frac{S_2}{P_2}$

The solubility of argon is S₁ at P₁ pressure and S₂ at pressure P₂.

From the graph we got, S₁ = 55 mg/100 gm water, P₁=10 atmand P₂ = 15 atm

$\begin{array}{l}\frac{S_1}{P_1}=\frac{S_2}{P_2}\\ Therefore,S_2=\frac{S_1\times P_2}{P_1}\\ =\frac{55mg/100gmwater\times 15atm}{10atm}=82.5mg/100gmwater\end{array}$

So, the solubility of argon at pressure 15 atm is 82.5 mg/100 gm water.