A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is ___________ g mol⁻¹. (Nearest integer)
Given, water boils at 373 K, K_b for water =0.52 K kg mol⁻¹

Solution

Chemistry ma’am · 4 steps

Here Δ T_b equals K_b, so the molality is exactly 1. Find moles from that, then the molar mass.

1. Find the rise. The solution boils at 373.52 K and pure water boils at 373 K. So Δ T_b = 0.52 K.

   Δ T_b = K_b m = K_b × (wₛₒₗᵤₜₑ/M)/(Wₛₒₗᵥₑₙₜ (kg))

2. Get the molality. m = (Δ T_b)/K_b = 0.52/0.52 = 1 mol kg⁻¹. So 1 kg of water would hold 1 mol of solute.
3. Moles in 20 g. The solvent is 20 g = 0.020 kg. Moles of solute = 1 × 0.020 = 0.02 mol.
4. So. M = (2 g)/(0.02 mol) = 100 g mol⁻¹, so the answer is 100.

The common slip: 0.1. Using the solvent mass as 20 instead of 0.020 kg gives 0.1, not 100. Convert grams of water to kilograms before you use the molality.

The idea to keep. Molality uses kilograms of solvent. Get m from Δ T_b / K_b, multiply by the solvent mass in kg to get moles, then divide the solute mass by those moles.

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