Question

$30mL$ of $C{H}_{3}OH(d=0.8g/c{m}^3)$ is mixed with $60mL$ of $C_2{H}_{5}OH(d=0.92g/c{m}^3)$ at ${25}^{\circ }C$ to form a solution of density $0.88g/c{m}^3$. Select the correct option.

• A. Molarity and molality of resulting solution are 6.33 and 13.59 respectively
• B. The mole fraction of solute and molality are 0.385 and 13.59 respectively
• C. Molarity and $\%$ change in volume are 13.59 and zero respectively
• D. Mole fraction of solvent and molality are 0.615 and 13.59 respectively

Answer 1

Answer: (B), (D)

The correct options are
B The mole fraction of solute and molality are 0.385 and 13.59 respectively
D Mole fraction of solvent and molality are 0.615 and 13.59 respectively

Number of moles of methanol $=\frac{30\times 0.8}{32}=0.75$

Number of moles of ethanol $=\frac{60\times 0.92}{46}=1.2$

Mole fraction of solute methanol $=\frac{0.75}{0.75+1.2}=0.385$

Mole fraction of solvent ethanol $=1-0.385=0.615$

Molality of solution $=\frac{0.75\times 1000}{1.2\times 46}=13.59m$

Thus, options B and D are correct.

Volume of the solution $=\frac{30\times 0.8+60\times 0.92}{0.88}=90c{m}^3$

The molarity of solution $=\frac{0.75\times 1000}{90}=8.33M$

Thus, options A and C are incorrect.