Question

Calculate the molar mass of nonvolatile solute when 4 g of it is dissolved in 100 g solvent that boils at 319.4 K [\(K_b\) = 2.4 K kg \(\text{mol}^{-1}\); B.P of pure solvent = 319 K]

• A. \(220 \text{gmol}^{-1}\)
• B. \(230 \text{gmol}^{-1}\)
• C. \(240 \text{gmol}^{-1}\)
• D. \(250 \text{gmol}^{-1}\)

Hint:

Use \(\Delta T_b = K_b \times \dfrac{w_2\times1000}{M_2\times w_1}\).

Answer 1

The Correct Option is C

Step 1: Find the molality first
$m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.4}{2.4} = \dfrac16$ mol per kg solvent.

Step 2: Convert to molar mass
100 g solvent is 0.1 kg, so moles of solute $= \frac16\times0.1 = \frac{1}{60}$ mol.
$M = \dfrac{4}{1/60} = 240$ g/mol, option (C).

Final Answer:
The solute has molar mass 240 g/mol, option (C). \[ \boxed{240\ \text{g mol}^{-1}} \]