1 g of non-volatile non-electrolyte solute is dissolved in 100g 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 points, (Δ Tb(A)/Δ Tb (B)), is:

Question

1 g of non-volatile non-electrolyte solute is dissolved in 100g 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 points, (Δ Tb(A)/Δ Tb (B)), is: (A) 5: 1 (B) 10: 1 (C) 1: 5 (D) 1: 0.2

Tardigrade Admin · Accepted Answer

Δ Tb = Kb × m ∴ (Δ Tb(A)/Δ Tb(B)) = (Kb(A)/Kb(B)) as mA =mB ∴ (Δ Tb(A)/Δ Tb(B)) = (1/5)

Q. 1 g of non-volatile non-electrolyte solute is dissolved in 100g 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 points, $\frac{Δ T _{b} ( A )}{Δ T _{b} ( B )}$ , is:

5 : 1

10 : 1

1 : 5

1 : 0.2

$Δ T_{b} = K_{b} \times m$
$∴ \frac{Δ T _{b (A)}}{Δ T _{b (B)}} = \frac{K _{b (A)}}{K _{b (B)}} a s m_{A} = m_{B}$
$∴ \frac{Δ T _{b (A)}}{Δ T _{b (B)}} = \frac{1}{5}$

Q. 1 g of non-volatile non-electrolyte solute is dissolved in 100g 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 points, ΔTb​(B)ΔTb​(A)​, is: