Question

When a liquid that is immiscible with water was steam distilled at ${95.2}^o$C at a total pressure of $99.652$ kPa, the distillate contained $1.27$ g 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 ${95.2}^o$C?

• A. $99.65gmo{l}^{-1}$
• B. $18gmo{l}^{-1}$
• C. $134.1gmo{l}^{-1}$
• D. $105.74gmo{l}^{-1}$

Answer 1

The correct option is C $134.1gmo{l}^{-1}$

Total pressure, $P_{total}=99.652$ kPa

$P_{water}=p_B=85.140$ kPa

$P_{liquid}=p_A=(99.652-85.140)$ 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, $M_A=\left(\frac{m_A}{m_B}\right)\left(\frac{p_B{M}_B}{p_A}\right)$

$M_A=1.27\times \left(\frac{85.140kPa\times 18gmo{l}^{-1}}{14.512kPa}\right)$

$M_A\simeq 134.1gmo{l}^{-1}$