Elevation in the boiling point for 1 molal solution of glucose is 2K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2K. The relation between Kb and Kf is:

Engineering

Chemistry

Colligative properties and Measurement

Elevation in Boiling Point

Depression in Freezing Point

Question

Elevation in the boiling point for 1 molal solution of glucose is 2K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2K. The relation between K_b and K_f is:

K_b = 1.5K_f

K_b = K_f

K_b = 0.5K_f

K_b = 2K_f

${ΔT}_{b} = K_{b} . m; {ΔT}_{f} = K_{f} . m$

2 = K_b.1 2 = K_f .2

⇒ K_b = 2.K_f

This question involves colligative properties, specifically boiling point elevation and freezing point depression. Let's break down the given information and solve step by step.

Step 1: Understand the formulas

The boiling point elevation (ΔT_b) is given by:

Δ T_{b} = K_{b} \times m

where K_b is the molal boiling point elevation constant and m is the molality.

The freezing point depression (ΔT_f) is given by:

Δ T_{f} = K_{f} \times m

where K_f is the molal freezing point depression constant and m is the molality.

Step 2: Apply the given data

For the 1 molal glucose solution:

Δ T_{b} = K_{b} \times 1 = 2 K

So, K_b = 2 K (since ΔT_b = 2 K for m=1).

For the 2 molal glucose solution:

Δ T_{f} = K_{f} \times 2 = 2 K

So, 2K_f = 2 K, which means K_f = 1 K.

Step 3: Relate K_b and K_f

From above, K_b = 2 K and K_f = 1 K. Therefore:

K_{b} = 2 K_{f}

Final Answer: K_b = 2K_f

Detailed Solution

Related Topics and Formulae

Detailed Solution

Related Topics and Formulae

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