Question

An aqueous solution containing 1.70 g of cane sugar in 100 mL water begins to freeze at −0.093°C. The cryoscopic constant (molal depression constant) of water is 1.86 K kg mo1⁻¹. Calculate the molecular weight of cane sugar.

Answer 1

Given: K_b = 1.86 K kg mo1⁻¹

Wt. of solute (water) = w = 1.70 g

Volume of water = Weight of water     ...(∵ density of water = 1 g/cc)

Weight of solvent (water) = W = 100 g

Freezing point of solution = −0.093°C

Depression of freezing point = ΔT_f

= Freezing point of water − Freezing point of solution

= 0 − (−0.093)

= 0.093°C

Molecular weight of cane sugar = M = ?

`M = (1000 xx K_f xx w)/(DeltaT_f xx W)`

= `(1000 xx 1.86 xx 1.70)/(0.093 xx 100)`

= 340 g/mol