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

The Correct Option is C

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.