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The molar mass of the compound is 125 g/mol. This result comes from applying the freezing point depression formula to the given data.

Problem Data

80 g acetic acid serves as the solvent with K_f = 3.9 K kg mol^-1. Adding 20 g of an unknown compound lowers the freezing point by ΔT_f = 7.8 K.

Solution Steps

The freezing point depression follows ΔT_f = K_f × m, where m is molality (moles solute per kg solvent).

• Molality m = ΔT_f/K_f = 7.8/3.9 = 2.0 mol kg^-1
• Mass of solvent = 80/1000 = 0.08 kg
• Moles of solute = m × 0.08 = 2.0 × 0.08 = 0.16 mol
• Molar mass = 20/0.16 = 125 g mol^-1 (rounded to nearest integer)

Direct Formula Method

Molar mass M = (w_B × K_f × 1000)/(ΔT_f × W_A)

Where w_B = solute mass (20g), W_A = solvent mass (80g)

M = (20 × 3.9 × 1000)/(7.8 × 80) = 125 g mol^-1

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🔬 CSIR NET Freezing Point Depression Guide

Introduction to Freezing Point Depression

Freezing point depression calculations determine unknown molar mass using colligative properties. In this CSIR NET-style problem, 80g acetic acid with freezing point constant 3.9 K kg mol⁻¹ experiences a 7.8K drop from 20g solute, yielding molar mass 125 g mol⁻¹.

Key Formula and Derivation

Use ΔT_f = K_f × m for non-electrolytes (van’t Hoff factor i=1). Molality m = moles solute/kg solvent.

Step-by-Step Calculation

1. Compute molality: m = 7.8/3.9 = 2 mol kg⁻¹
2. Solvent in kg: 80/1000 = 0.08 kg
3. Moles solute: 2 × 0.08 = 0.16 mol
4. Molar mass: 20/0.16 = 125 g mol⁻¹

CSIR NET Exam Tips

• Practice assumes non-electrolyte behavior (i=1)
• Acetic acid’s K_f=3.9 is standard
• Common errors: forgetting kg conversion, rounding prematurely
• Verify with similar problems like benzoic acid in acetic acid