Question

4.6 cm${}^3$ of methyl alcohol is dissolved in 25.2 g of water. Calculate
(i) % by mass of methl alcohol
(ii) mole fraction of methyl alcohol and water (Given density of methyl alcohol = 0.7952 gcm${}^{-3}$ and C = 12, H = 1, O = 16)
i) 12.68 ii) 0.0755, 0.9245

Answer 1

Volume of methyl alcohol $=4.6{cm}^3$

Density of methyl alcohol $=0.7952g{cm}^{-3}$

$\Rightarrow Density=\frac{Mass}{Volume}$

$\Rightarrow 0.7952=\frac{Mass}{4.6}$

$\Rightarrow Mass=3.657g$

$\therefore$ Mass of methyl alcohol $=3.657g$

(i) $\%$ by mass of methyl alcohol $=\frac{\text{mass of methyl alcohol}}{\text{mass of methyl alcohol}+\text{mass of water}}\times 100=\frac{3.657}{3.657+25.2}\times 100=12.6\%$

(ii) Mole fraction of methyl alcohol and water-

Moles of methyl alcohol $=\frac{\text{Mass of methyl alcohol}}{\text{Molecular mass of methyl alcohol}}$

Molecular mass of methyl alcohol $\left(C{H}_{3}OH\right)=12+4\times 1+16=32g$

$\therefore$ Moles of methyl alcohol $=\frac{3.657}{32}=0.114$

Moles of water $=\frac{\text{Mass of water}}{\text{Molecular mass of water}}=\frac{25.6}{18}=1.42$

Total no. of moles = moles of methyl alcohol + moles of water $=0.114+1.42=1.534$

$\therefore$ Mole fraction of methyl alcohol $=\frac{\text{moles of methyl alcohol}}{\text{total no. of moles}}=\frac{0.114}{1.534}=0.075$

Mole fraction of water $=\frac{\text{moles of water}}{\text{total no. of moles}}=\frac{1.42}{1.534}=0.925$