Question

When a liquid that is immiscible with water was steam distilled at $95.2^{\circ}C$ at a total pressure of $99.652\, KPa$. The distillate contained $1.27\, gm$ of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is $85.140\, KPa$ at $95.2^{\circ}C$ ?

• A. $134.1\, gm\, mol^{-1}$
• B. $105.74\, gm\, mol^{-1}$
• C. $99.65\, gm\, mol^{-1}$
• D. $18\, gm\, mol^{-1}$

Answer 1

Answer: (A)

Total pressure, $p_{\text {total }}=99.652\, kPa$

$p_{B} =p_{\text {water }}=85.140\, kPa$

$p_{\text {liquid }} =p_{A}=(99.652-85.140) kPa$

$=14.512\, kPa$

and $\frac{m_{A}}{m_{B}}= \frac{1.27}{1}$

or $\frac{m_{A}}{m_{B}}=\frac{p_{A} M_{A}}{p_{B} M_{B}}$

or, $M_{A} =\left(\frac{m_{A}}{m_{B}}\right)\left(\frac{p_{B} M_{B}}{p_{A}}\right)$

$=1.27 \times\left(\frac{85.140 kPa \times 18 g mol ^{-1}}{14.512 kPa }\right)$

$\simeq 134.1\, g\, mol ^{-1}$