Question

A $5.2$ molal aqueous solution of methyl alcohol, $C{H}_{3}OH$ is supplied. What is the mole fraction of methyl alcohol in the solution?

• A. $1.100$
• B. $0.090$
• C. $0.086$
• D. $0.050$

Answer 1

Answer: (C)

Molality $=\frac{\text{ Moles of solute }}{\text{ Mass of solvent (in }kg\text{ ) }}$
$=\frac{5.2molC{H}_{3}OH}{1kg{H}_{2}O}[1kg=1000g]$
$n_1\left(C{H}_{3}OH\right)=5.2,n_2\left(H_{2}O\right)=\frac{1000}{18}=55.56$
$\therefore n_1+n_2=5.20+55.56=60.76mol$
$\therefore x_{C{H}_{3}OH}=\frac{n_1}{n_1+n_2}=\frac{5.2}{60.76}=0.086$