The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of Kb and Kf i.e., Kb / Kf is 1: X. The value of X is [nearest integer]

Question

The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of Kb and Kf i.e., Kb / Kf is 1: X. The value of X is [nearest integer] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T b = iK b m 1 Δ T f = iK f m 2 (Δ T b /Δ T f )=( K b × 1/ K f × 2) ⇒ (3/6)=(1/2)=( K b / K f ) × (1/2) ( K b / K f )=(1/1) ⇒ x =1

Q. The elevation in boiling point for 1 molal solution of non-volatile solute $A$ is $3 K$ . The depression in freezing point for 2 molal solution of $A$ in the same solvent is $6 K$ . The ratio of $K_{b}$ and $K_{f}$ i.e., $K_{b} / K_{f}$ is $1 : X$ . The value of $X$ is [nearest integer]

$Δ T_{b} = i K_{b} m_{1} Δ T_{f} = i K_{f} m_{2}$
$\frac{Δ T _{b}}{Δ T _{f}} = \frac{K _{b} \times 1}{K _{f} \times 2}$
$\Rightarrow \frac{3}{6} = \frac{1}{2} = \frac{K _{b}}{K _{f}} \times \frac{1}{2}$
$\frac{K _{b}}{K _{f}} = \frac{1}{1} \Rightarrow x = 1$

Q. The elevation in boiling point for 1 molal solution of non-volatile solute A is 3K. The depression in freezing point for 2 molal solution of A in the same solvent is 6K. The ratio of Kb​ and Kf​ i.e., Kb​/Kf​ is 1:X. The value of X is [nearest integer]

ΔTb​=iKb​m1​ΔTf​=iKf​m2​ ΔTf​ΔTb​​=Kf​×2Kb​×1​ ⇒63​=21​=Kf​Kb​​×21​ Kf​Kb​​=11​⇒x=1

Q. The elevation in boiling point for 1 molal solution of non-volatile solute $A$ is $3 K$ . The depression in freezing point for 2 molal solution of $A$ in the same solvent is $6 K$ . The ratio of $K_{b}$ and $K_{f}$ i.e., $K_{b} / K_{f}$ is $1 : X$ . The value of $X$ is [nearest integer]

$Δ T_{b} = i K_{b} m_{1} Δ T_{f} = i K_{f} m_{2}$
$\frac{Δ T _{b}}{Δ T _{f}} = \frac{K _{b} \times 1}{K _{f} \times 2}$
$\Rightarrow \frac{3}{6} = \frac{1}{2} = \frac{K _{b}}{K _{f}} \times \frac{1}{2}$
$\frac{K _{b}}{K _{f}} = \frac{1}{1} \Rightarrow x = 1$