Q. A 5.2 molal aqueous solution of methyl alcohol, ${CH}_{3}OH$, is supplied. What is the mole fraction of methyl alcohol in the solution?
A) 1.100
B) 0.190
C) 0.086
D) 0.050
Explanation
(c)
Molality $\frac{Moles\ of\ solute\ }{kg\ of\ solvent}$= $\frac{5.2\ mol\ {CH}_{3}OH}{kg\ ( = 100g)H_{2}O}$
$n_{1}\left( {CH}_{3}OH \right) = 5.2$
$\ \ \ \ \ \ \ n_{2}\left( H_{2}O \right) = \frac{1000}{18} = 55.56$
${\therefore n}_{1}{+ n}_{2}$ = 5.20 + 55.56 = 60.76 mol
${\therefore X}_{{CH}_{3}OH} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{5.2}{60.76} = 0.086$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.