Question

The mole fraction of $C{H}_{3}OH$ in a solution containing $64$g of $C{H}_{3}OH$ and $144$ml of $H_{2}O$ at $4^0$C is:

• A. $0.05$
• B. $0.2$
• C. $0.4$
• D. $0.8$

Answer 1

Answer: (B)

$0.2$

Molecular weight of $C{H}_{3}OH$ is $32g$. Given weight is $64g$ .

Number of moles of $C{H}_{3}OH=\frac{\text{Given weight}}{\text{Molecular weight}}=\frac{64}{32}=2$

Given volume of water= $144ml$

Density= $\frac{\text{Mass}}{\text{volume}}$

Density of water is $1g/ml$

$1=\frac{Mass}{144}\Rightarrow$ Mass of water= $144g$

Number of moles of $H_{2}O=\frac{\text{Obtained weight}}{GMW}=\frac{144}{18}=8$ .

Mole fraction of $C{H}_{3}OH=\frac{C{H}_{3}OH\text{Moles}}{C{H}_{3}OH\text{Moles}+H_{2}O\text{Moles}}=\frac{2}{2+8}=\frac{2}{10}=0.2$