Question

$1 \,g$ methanol $\left( d =0.75\, g \,mL ^{-1}\right)$ is mixed with $1 \,g$ water $\left( d =1.00 g\, mL ^{-1}\right)$ at $298 \,K$. Thus, volume $\%$ of methanol $( V / V )$ is

• A. $42.86 \%$
• B. $57.14 \%$
• C. $50.00 \%$
• D. $25.00 \%$

Answer 1

Answer: (B)

Methanol $=1 \,g =\frac{1}{0.75} mL =\frac{100}{75} mL$

$H _{2} O =1\, g =\frac{1}{1} mL =\frac{75}{75} mL$

Total volume of mixture $=\frac{100}{75}+\frac{75}{75}=\frac{175}{75}$

$\therefore \%$ methanol $=\left(\frac{\frac{100}{75}}{\frac{175}{75}}\right) \times 100=\frac{400}{7}=57.14 \%$