JEE MAIN - Chemistry (2019 - 10th January Evening Slot - No. 15)
Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is
Kb = Kf
Kb = 0.5 Kf
Kb = 1.5 Kf
Kb = 2 Kf
Explanation
$${{\Delta {T_b}} \over {\Delta {T_f}}} = {{i.m \times {k_b}} \over {i \times m \times {k_f}}}$$
$${2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}$$
$${k_b} = 2{k_f}$$
$${2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}$$
$${k_b} = 2{k_f}$$
Comments (0)
