A solution containing 2.5 × 10⁻³ kg of a solute dissolved in 75 × 10⁻³ kg of water boils at 373.535 K. The molar mass of the solute is ____________ g mol⁻¹. [nearest integer] (Given : K b (H₂O) = 0.52 K kg mol⁻¹ and boiling point of water = 373.15 K)

Solution

Chemistry ma’am · 4 steps

Get Δ T_b from the two boiling points, turn it into molality, then into moles and molar mass.

1. Find the rise. Δ T_b = 373.535 - 373.15 = 0.385 K.

   Δ T_b = K_b m = K_b × (w_B / M_B)/(W_A (in kg))

2. Molality. m = 0.385/0.52 = 0.7404 mol kg⁻¹. With 0.075 kg of water, the moles of solute are 0.7404 × 0.075 = 0.05553 mol.
3. Molar mass. The solute mass is 2.5 g, so M_B = 2.5/0.05553 = 45.02 g mol⁻¹.
4. So. The molar mass of the solute is about 45.02 g mol⁻¹, which rounds to 45.

The common slip: 0.045. Keeping the solute mass in kg (2.5 × 10⁻³ kg) gives the molar mass in kg mol⁻¹, which is 0.045, not 45.

The idea to keep. For molar mass from Δ T_b, use M_B = (K_b × w_B × 1000)/(Δ T_b × W_A) with both masses in grams. Watch the units so the answer comes out in g mol⁻¹.

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