When 3.00 g of a substance 'X' is dissolved in 100 g of CCl₄, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ______________ g mol⁻¹. (Nearest integer).
[Given K b for CCl₄ is 5.0 K kg mol⁻¹]

Solution

Chemistry ma’am · 3 steps

The rise in boiling point gives you the molality, and molality gives you the molar mass.

1. Set up molality. Δ T_b = K_b m, where m is moles of solute per kilogram of solvent. Moles of X are w/M, and the solvent mass in kg is W/1000.

   M = (K_b × w × 1000)/(Δ T_b × W)

2. Put in numbers. w = 3.00 g, W = 100 g, K_b = 5.0 K kg mol⁻¹, Δ T_b = 0.60 K. So M = (5.0 × 3.00 × 1000)/(0.60 × 100) = 15000/60.
3. So. M = 250 g mol⁻¹, so the answer is 250.

The common slip: 0.25. Leaving the solvent mass as 100 instead of 0.1 kg gives 0.25, not 250. Divide the grams of solvent by 1000.

The idea to keep. For molar mass from any colligative property, find the molality first. Always put the solvent mass in kilograms, which is where the factor of 1000 comes from.

Answer from the NTA final answer key. Solution by PrepWiser.