Question

Calculate the mole percentage of $C{H}_{3}OH$ and $H_{2}O$ respectively in $60$% (by mass) aqueous solution of $C{H}_{3}OH$.

• A. $45.8,54.2$
• B. $54.2,45.8$
• C. $50,50$
• D. $60,40$

Answer 1

Answer: (A)

$45.8,54.2$

Given,

$60$% by mass solution of $C{H}_{3}OH$. That means in $100gm$ solution

$60gm$ $C{H}_{3}OH$ is dissolved in $40gm$ $H_{2}O$

$\therefore$ Molar mass of $C{H}_{3}OH=32g/mol$

Molar mass of $H_{2}O=18g/mol$

$\Rightarrow$ No. of moles of $C{H}_{3}OH=\frac{60gm}{32g/mol}=1.875mol$ $..........\left(1\right)$

No. of moles of $H_{2}O=\frac{40gm}{18gm/mol}=2.222mol$ $..........\left(2\right)$

Total no. of moles=$1.875+2.222=4.097$ moles $..........\left(3\right)$

$\Rightarrow$ Mole percentage= $\frac{Molesofcompound}{Totalno.ofmoles}\times 100$

Mole % of $C{H}_{3}OH=\frac{1.875}{4.097}\times 100$

$=45.8$% (From 1 & 3)

Mole % of $H_{2}O=\frac{2.222}{4.097}\times 100$

$=54.23$% (from 2,3)

Hence, the correct option is $A$