Question

What is the molar mass of solute when 2.3 gram non-volatile solute dissolved in 46 gram benzene at 30°C? (Relative lowering of vapour pressure is 0.06 and molar mass of benzene is 78 gram mol–1)

• A. 72 gram mol–1
• B. 48 gram mol–1
• C. 65 gram mol–1
• D. 80 gram mol–1

Answer 1

Answer: (C)

65 gram mol–1

Answer 2

To find the molar mass of the solute, we can use the formula for relative lowering of vapor pressure:
$\frac {ΔP}{P_₀}$= $\frac {n_2}{n_1}$ . $\frac {M_1}{M_2}$
Where
$\frac {ΔP}{P_₀}$ is the relative lowering of vapor pressure,
n2 is the number of moles of solute,
n1 is the number of moles of solvent,
M1 is the molar mass of the solvent,
M2 is the molar mass of the solute.
Given:
$\frac {ΔP}{P_₀}$ = 0.06
n₂ = 2.3 g
n₁ = 46 g
M1 = 78 g/mol (molar mass of benzene)
M₂ = ?
Substituting the given values into the formula, we get:
0.06 = $\frac {2.3 \times 78}{46 \times M₂}$
0.06 x (46 x M₂) = 2.3 x 78
2.76 x M₂ = 179.4
M₂ = $\frac {179.4}{2.76}$
M₂ ≈ 65 gram/mol
Therefore, the molar mass of the solute is approximately 65 gram/mol.
The correct answer is option (C) 65 gram/mol.