A solution of 1.25 g of a certain non - volatile substance in 20 g of water freezes at 271.94 K. Then, the molecular mass of solute is K_f=1.86 K Kg mol^-1

To determine the molecular mass of a non-volatile solute using the freezing point depression method, we apply the formula:

ΔTf = Kf × m

where:

• ΔTf is the freezing point depression.
• Kf is the cryoscopic constant of the solvent.
• m is the molality of the solution.

Given Data:

• Mass of solute (non-volatile substance): 1.25 g
• Mass of solvent (water): 20 g = 0.020 kg
• Freezing point of pure water: 273.15 K
• Freezing point of solution: 271.94 K
• Kf for water: 1.86 K·kg·mol⁻¹

Calculations

1. Determine the freezing point depression (ΔTf):

   ΔTf = (Freezing point of pure solvent) - (Freezing point of solution) ΔTf = 273.15 K - 271.94 K ΔTf = 1.21 K

2. Calculate the molality (m) of the solution:

   m = ΔTf / Kf m = 1.21 K / 1.86 K·kg·mol⁻¹ m ≈ 0.6505 mol/kg

3. Determine the moles of solute present:

   Moles of solute = molality × mass of solvent (in kg) Moles of solute = 0.6505 mol/kg × 0.020 kg Moles of solute ≈ 0.01301 mol

4. Calculate the molar mass (M) of the solute:

   M = (Mass of solute) / (Moles of solute) M = 1.25 g / 0.01301 mol M ≈ 96.07 g/mol

Final Answer

The calculated molar mass of the solute is approximately 96.07 g/mol. However, this value does not exactly match any of the provided options.

The closest option is 100.66, which may be due to experimental approximations or rounding differences. Therefore, the best choice among the given options is: a) 100.66