Question

A solution of sucrose (molar mass = 342 g mol−1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be : (Kf for water = 1.86 K kg mol –1)

Answer 1

weight of sucrose = 68.5 g

molecular weight of sucrose = 342 g/mol

so, no of moles of sucrose = 68.5/342 = 0.2 mol

mass of solvent (water) = 1000g

molality = no of moles of sucrose/mas of solvent in kg

= 0.2/(1000g/1000)

= 0.2molal

using formula, ∆Tf = kf × molality

here, kf = 1.86 Kg.°C/mol and molality = 0.2

so, ∆Tf = 1.86 × 0.2 = 0.372°C

hence, the freezing point of the solution obtained will be 0.372°C

Answer 2

Answer:

Explanation:

Refer the below attachment.