Question

A mixture of ethanol and propanol has vapor pressure 279 mm of Hg at 300 K. The mol fraction of ethanol in the solution is 0.6 and vapour pressure of pure propanol is 210 mm of Hg. Vapour pressure of pure ethanol will be:

• A. $336.6$ mm
• B. $325$ mm
• C. $375$ mm
• D. $415$ mm

Answer 1

Answer: (B)

$325$ mm

$p=p_1+p_2$

$\Rightarrow p={p_1}^0{x}_1+{p_2}^0{x}_2$ ( Raoult's law )

$p={p_1}^0(1-x_2)+{p_2}^0{x}_2$

$p={p_1}^0+({p_2}^0-{p_1}^0)x_2$

$279mm={p_1}^0+({p_2}^0-{p_1}^0)x_2$

Let $2,$ ethanol

$1,$ propanol

$\therefore 279mm=210mm+({p_2}^0-210)0.6$

$69=({p_2}^0-210)0.6$

${p_2}^0-210=115$

$\Rightarrow {p_2}^0=325mm$

Hence, the answer is $325mm.$