Question

The molal freezing point constant for water is 1.86. If 342g of canesugar (C12H22O11) are dissolved in 1000g water the solution will freeze at

Answer 1

delta(T_f)=-K_f m
molality of solution =(342/342)/1

delta(T_f)=-1.86×(342/342)/1=-1.86°c

so the solution will freeze at -1.86°c

Answer 2

Answer: 271.14K

Explanation:

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

where,

$\Delta T_f=T^{o}_f-T_f$

$T^{o}_f$= freezing point of pure solvent i.e water

$T_f$= freezing point of solution

$273K-T_f$ = change in freezing point

$k_f$ = freezing point constant = 1.86 K kg/mol

m = molality

$\Delta T_f=k_f\times \frac{\text{mass of sugar}}{\text{molar mass of sugar}\times \text{weight of solvent in kg}}$

$273-T_f=1.86\times \frac{342 g}{342g/mol\times 1kg}$

$T_f=271.14K$