Solute ‘X’ dimerises in water to the extent of 80 %. 2.5 g of ‘X’ in 100 g of water increases the boiling point by 0.3 ° C. The molar mass of ‘X’ is [Kb=0.52 K kg mol-1]

Question

Solute ‘X’ dimerises in water to the extent of 80 %. 2.5 g of ‘X’ in 100 g of water increases the boiling point by 0.3 ° C. The molar mass of ‘X’ is [Kb=0.52 K kg mol-1] (A) 13 (B) 52 (C) 65 (D) 26

Tardigrade Admin · Accepted Answer

We are given, 2 X arrow X 2 Therefore, i =1-α+ propto / 2=1-0.8+0.4=0.6 (since α=0.8, given ) We know, Δ T b = ik b × W B / M × 1000 /( W A ( g )) Where, W B = Mass of solute, W A = Mass of solvent, M = Molar mass Given, Δ T b =0.3 Therefore, 0.3=0.6 × 0.52 × 2.5 / M × 1000 / 100 ⇒ M =2 × 0.52 × 2.5 × 10=26 g

Q. Solute ‘X’ dimerises in water to the extent of 80%.2.5g of ‘X’ in 100g of water increases the boiling point by 0.3∘C. The molar mass of ‘X’ is [Kb​=0.52Kkgmol−1]

13

52

65

26

We are given, 2X→X2​ Therefore, i=1−α+∝/2=1−0.8+0.4=0.6 (since α=0.8, given ) We know, ΔTb​=ikb​×WB​/M×1000/(WA​(g)) Where, WB​= Mass of solute, WA​= Mass of solvent, M= Molar mass Given, ΔTb​=0.3 Therefore, 0.3=0.6×0.52×2.5/M×1000/100 ⇒M=2×0.52×2.5×10=26g

Q. Solute ‘ $X$ ’ dimerises in water to the extent of $80%.2.5 g$ of ‘ $X$ ’ in $100 g$ of water increases the boiling point by $0. 3^{\circ} C$ . The molar mass of ‘ $X$ ’ is $[K_{b} = 0.52 K k g m o l^{- 1}]$