The elevation in boiling point of an aqueous solution of NaCl is 0.01° C. If its van't Hoff factor is 1.92, the molality of NaCl solution is (Kb. for water =0.52 K kg mol -1 )

Question

The elevation in boiling point of an aqueous solution of NaCl is 0.01° C. If its van't Hoff factor is 1.92, the molality of NaCl solution is (Kb. for water =0.52 K kg mol -1 ) (A) 0.01 m (B) 0.001 m (C) 0.005 m (D) 0.02 m

Tardigrade Admin · Accepted Answer

Given, Δ Tb=0.01° C =0.01 K i =1.92 Kb =0.52 K kg mol -1 As we know that, Δ Tb =i Kb × m ∴ m =(0.01/1.92 × 0.52) mol / kg m (molality) =0.01 mol / kg or 0.01 m

Q. The elevation in boiling point of an aqueous solution of $N a Cl$ is $0.0 1^{\circ} C$ . If its van't Hoff factor is $1.92$ , the molality of $N a Cl$ solution is
$(K_{b}$ for water $= 0.52 K k g m o l^{- 1}$ )

0.01 m

0.001 m

0.005 m

0.02 m

Given, $Δ T_{b} = 0.0 1^{\circ} C = 0.01 K$

$i = 1.92$

$K_{b} = 0.52 K k g m o l^{- 1}$

As we know that,

$Δ T_{b} = i K_{b} \times m$

$∴ m = \frac{0.01}{1.92 \times 0.52} m o l / k g$

$m$ (molality) $= 0.01 m o l / k g$ or $0.01 m$

Q. The elevation in boiling point of an aqueous solution of NaCl is 0.01∘C. If its van't Hoff factor is 1.92, the molality of NaCl solution is (Kb​ for water =0.52Kkgmol−1 )