Question

A 3.5 molal aqueous solution of methyl alcohol $(CH_{3}OH)$ is supplied. What is the mole fraction of methyl alcohol in the solution?

• A. $0.100$
• B. $0.059$
• C. $0.086$
• D. $0.050$

Answer 1

Answer: (B)

$\frac{\text{Moles of(solute) $CH_{3}$ OH}}{\text{1kg of $H_{2}$O}} = \frac{3.5 mol}{1kg}$
Thus, $n\left(CH_{3}OH\right)=3.5$
$N\left(H_{2}O\right)=\frac{1000}{18}=55.56$
$\therefore $ Total moles $=3.5+55.56=59.06\, mol$
$\therefore \chi_{CH_3OH} =\frac{\text{Moles of $CH_{3}$OH}}{\text{Total moles}}$
$=\frac{3.5}{59.05}=0.059$