1g of non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A textandB whose ebullioscopic constants are in the ratio of 1:5 . The ratio of the elevation in their boiling point, (Δ Tb (A)/Δ Tb (B)), is

Question

1g of non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A textandB whose ebullioscopic constants are in the ratio of 1:5 . The ratio of the elevation in their boiling point, (Δ Tb (A)/Δ Tb (B)), is (A) 5:1 (B) 10:1 (C) 1:5 (D) 1:3

Tardigrade Admin · Accepted Answer

Δ Tb=Kb× m⇒ Δ Tb propto Kb As mA=mB (given) So (Δ Tb (A)/Δ Tb (B))=(Kb (A)/Kb (B)) As (Kb (A)/Kb (B))=(1/5) So (Δ Tb (A)/Δ Tb (B))=(1/5)

Q. $1 g$ of non-volatile non-electrolyte solute is dissolved in $100 g$ of two different solvents $A and B$ whose ebullioscopic constants are in the ratio of $1 : 5$ . The ratio of the elevation in their boiling point, $\frac{Δ T _{b} ( A )}{Δ T _{b} ( B )},$ is

$5 : 1$

$10 : 1$

$1 : 5$

$1 : 3$

$Δ T_{b} = K_{b} \times m \Rightarrow Δ T_{b} \propto K_{b}$
As $m_{A} = m_{B}$ (given)
So $\frac{Δ T _{b (A)}}{Δ T _{b (B)}} = \frac{K _{b (A)}}{K _{b (B)}}$
As $\frac{K _{b (A)}}{K _{b (B)}} = \frac{1}{5}$
So $\frac{Δ T _{b (A)}}{Δ T _{b (B)}} = \frac{1}{5}$

Q. 1g of non-volatile non-electrolyte solute is dissolved in 100g of two different solvents AandB whose ebullioscopic constants are in the ratio of 1:5 . The ratio of the elevation in their boiling point, ΔTb​(B)ΔTb​(A)​, is

5:1

10:1

1:5

1:3