Question

If $p_A^o$ is the vapour of pure solvent and $p_A$ is the vapour pressure of solution. If $X_B$ is the mole fraction of solute and $X_A$ is the mole fraction of the solvent. What is the expression for relative lowering of vapour pressure of the solution?

• A. $\frac{p_A^o-p_A}{p_A^o}$$=$$X_B$
• B. $\frac{p_A^o-p_A}{p_A^o}$$=$$X_A$
• C. $\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{n_B}{n_A}$
• D. $\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{n_A}{n_B}$

Answer 1

Answer: (A), (C)

$\frac{p_A^o-p_A}{p_A^o}$$=$$X_B$
$\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{n_B}{n_A}$

Relatieve lowering of vapour pressure of a solution is a colligative property which depends only on the number of solute particles. So relative lowering of vapour pressure depends on the mole fraction of the solute too. So the correct expression for the relative lowering of vapour pressure of a solution is $\frac{p_A^o-p_A}{p_A^o}$$=$$X_B$
$X_B$ that is the mole fraction of solute can also be written as $\frac{n_B}{n_A+n_B}$, where $n_B$ and $n_A$ are the number of moles of solute and solvent. But the amount of solute present is very less compared to the amount of solvent and so $n_B$ in the denominator can be removed. Thus the equation can also be written as $\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{n_B}{n_A}$.