1.80 g of solute A was dissolved in 62.5 cm³ of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is ________ g mol⁻¹.
[Given : Freezing point of ethanol is 156.0 K.
Density of ethanol is 0.80 g cm⁻³.
Freezing point depression constant of ethanol is 2.00 K kg mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Turn the ethanol volume into a mass first, then let the freezing-point drop give you the molality.

1. Mass of solvent. The solvent is given as a volume, but molality needs its mass. Mass of ethanol = 62.5 cm³ × 0.80 g cm⁻³ = 50 g = 0.050 kg.
2. Depression. Δ T_f = 156.0 - 155.1 = 0.9 K.

   Δ T_f = K_f × (w_A / M_A)/(Wₛₒₗᵥₑₙₜ (kg))

3. Solve for M. M_A = (K_f × w_A)/(Δ T_f × W) = (2.00 × 1.80)/(0.9 × 0.050) = 3.6/0.045 = 80 g mol⁻¹.
4. So. The molar mass of solute A is 80 g mol⁻¹.

The common slip: 64. Using 62.5 cm³ directly as 62.5 g of ethanol, without the density, gives 64.

The idea to keep. Molality always needs the mass of the solvent in kilograms. When the solvent is given as a volume, multiply by its density before you do anything else.

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