The boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1° C higher than that of pure ether. The molecular weight of the substance will be (Kb=2.16° C kg mol -1)

Question

The boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1° C higher than that of pure ether. The molecular weight of the substance will be (Kb=2.16° C kg mol -1) (A) 148 (B) 158 (C) 168 (D) 178

Tardigrade Admin · Accepted Answer

Δ Tb=(w2 × 1000 × Kb/M2 × w1) M2=(1000 Kb w2/w1 Δ Tb)=(1000 × 2.16 × 0.11/15 × 0.1) =158.4 ≈ 158

Q. The boiling point of a solution of $0.11 g$ of a substance in $15 g$ of ether was found to be $0. 1^{\circ} C$ higher than that of pure ether. The molecular weight of the substance will be $(K_{b} = 2.1 6^{\circ} C k g m o l^{- 1})$

148

158

168

178

$Δ T_{b} = \frac{w _{2} \times 1000 \times K _{b}}{M _{2} \times w _{1}}$

$M_{2} = \frac{1000 K _{b} w _{2}}{w _{1} Δ T _{b}} = \frac{1000 \times 2.16 \times 0.11}{15 \times 0.1}$

$= 158.4 \approx 158$

Q. The boiling point of a solution of 0.11g of a substance in 15g of ether was found to be 0.1∘C higher than that of pure ether. The molecular weight of the substance will be (Kb​=2.16∘Ckgmol−1)

148

158

168

178

ΔTb​=M2​×w1​w2​×1000×Kb​​ M2​=w1​ΔTb​1000Kb​w2​​=15×0.11000×2.16×0.11​ =158.4≈158

Q. The boiling point of a solution of $0.11 g$ of a substance in $15 g$ of ether was found to be $0. 1^{\circ} C$ higher than that of pure ether. The molecular weight of the substance will be $(K_{b} = 2.1 6^{\circ} C k g m o l^{- 1})$

$Δ T_{b} = \frac{w _{2} \times 1000 \times K _{b}}{M _{2} \times w _{1}}$

$M_{2} = \frac{1000 K _{b} w _{2}}{w _{1} Δ T _{b}} = \frac{1000 \times 2.16 \times 0.11}{15 \times 0.1}$

$= 158.4 \approx 158$