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

Question

Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point of 2 molal solutions of glucose in the same solvent is 2 K. The relation between Kb and Kf is: (A) Kb = 0.5 Kf (B) Kb = 2 Kf (C) Kb = 1.5 Kf (D) Kb = Kf

Tardigrade Admin · Accepted Answer

(Δ Tb/Δ Tf) = (i.m × kb/i× m× kf) (2/2) = (1×1× kb/1 ×2× kf) kb = 2Kf

Q. Elevation in the boiling point for $1$ molal solution of glucose is $2 K$ . The depression in the freezing point of $2$ molal solutions of glucose in the same solvent is $2 K$ . The relation between $K_{b}$ and $K_{f}$ is:

$K_{b} = 0.5 K_{f}$

$K_{b} = 2 K_{f}$

$K_{b} = 1.5 K_{f}$

$K_{b} = K_{f}$

$\frac{Δ T _{b}}{Δ T _{f}} = \frac{i . m \times k _{b}}{i \times m \times k _{f}}$
$\frac{2}{2} = \frac{1 \times 1 \times k _{b}}{1 \times 2 \times k _{f}}$
$k_{b} = 2 K_{f}$

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

Kb​=0.5Kf​

Kb​=2Kf​

Kb​=1.5Kf​

Kb​=Kf​