Question

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]

Answer 1

Correct Answer: 80

To find the molar mass of solute A, we will use the formula for freezing point depression: ΔT_f = iK_fm, where ΔT_f is the change in freezing point, K_f is the freezing point depression constant, and m is the molality of the solution. Since solute A is non-electrolyte, i=1.

1. Calculate ΔT_f: ΔT_f = 156.0 K - 155.1 K = 0.9 K.

2. Determine the mass of ethanol in kg: Volume = 62.5 cm³, Density = 0.80 g/cm³.
Mass = Density × Volume = 0.80 g/cm³ × 62.5 cm³ = 50 g = 0.050 kg.

3. Calculate molality (m): m = n_solute/mass_solvent (kg). We rearrange ΔT_f = K_f m to find m = ΔT_f/K_f = 0.9 K / 2.00 K kg mol^–1 = 0.45 mol/kg.

4. Express molality as n_solute/0.050 kg = 0.45 mol/kg, yielding n_solute = 0.0225 mol.

5. Calculate molar mass (M) of solute A: M = mass_solute/n_solute = 1.80 g / 0.0225 mol = 80 g/mol.

6. Verification: The calculated molar mass (80 g/mol) lies within the given range (80, 80). Thus, the molar mass of solute A is confirmed to be 80 g/mol.

Answer 2

\(ΔT_f=k_fm\)

\(0.9=\frac {2×1.8×1000}{62.5×0.8×M}\)

Then, the molar mass of solute A,

\(M=\frac {2×1800}{62.5×0.8×0.9}\)

\(M=80\) g/mol

So, the answer is \(80\).