Question

From relative lowering of vapour pressure , molecular mass of non-volatile solute can be determined.
Which of the following relations is correct?
Here,
$P_A^{\circ }\text{is vapour pressure of pure solvent}$
$P_S\text{is vapour pressure of solution}$
$w_A,w_B\text{are mass of solvent and solute respectively}$
$M_A,M_B\text{are molecular masses of solvent and solute respectively}$

• A. $M_B=\frac{w_B\times M_A\times P_S}{w_A\times (P_A^{\circ }-P_S)}$
• B. $M_B=\frac{w_A\times M_A\times P_S}{w_B\times (P_A^{\circ }-P_S)}$
• C. $M_B=\frac{w_B\times (P_A^{\circ }-P_S)}{w_A\times M_A\times P_S}$
• D. None of the above

Answer 1

The correct option is A $M_B=\frac{w_B\times M_A\times P_S}{w_A\times (P_A^{\circ }-P_S)}$

Relative lowering of vapour pressure
When the solution is not dilute
$\frac{(P_A^{\circ }-P_S)}{P_A^{\circ }}=\frac{n_B}{n_A+n_B}$
On reversing the equation:
$\frac{P_A^{\circ }}{P_A^{\circ }-P_S}=1+\frac{n_A}{n_B}$

$\frac{P_A^{\circ }}{P_A^{\circ }-P_S}-1=\frac{n_A}{n_B}$

$\frac{P_S}{P_A^{\circ }-P_S}=\frac{n_A}{n_B}$

$\frac{P_A^{\circ }-P_S}{P_S}=\frac{n_B}{n_A}$

$\Rightarrow \frac{P_A^{\circ }-P_S}{P_S}=\frac{w_B\times M_A}{w_A\times M_B}$

$M_B=\frac{w_B\times M_A\times P_S}{w_A\times (P_A^{\circ }-P_S)}$