Question

calculate the freezing point of a solution containing 18g of glucose(C6H12O6) and 68.4 g of sucrose(C12H22O11) in 200 g of water.The freezing point of water is 273 K and Kf of water is 1.86 K.Kgmol-1

Answer 1

Answer:The freezing point of the solution is 271.5 K

Explanation:

Given mass of glucose = 18 g

Given mass of sucrose = 68.4 g

Molar mass of glucose = 180 g/mol

Molar mass of sucrose = 342 g/mol

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

${\text {Moles of glucose}=\frac{18g}{180g/mol}=0.1$

${\text {Moles of sucrose}=\frac{68.4g}}{342g/mol}=0.2$

Total moles = 0.1  + 0.2= 0.3 moles

As depression in freezing point is a colligative property, it only depends on the amount of the solute.

$\Delta T_f=k_f\times m$

$T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

$\Delta T_f=K_f\times \frac{\text{total moles of solute}}{\text {weight of solvent in kg}}$

$\Delta T_f=1.86Kkg/mol\frac{0.3}{0.2kg}=1.5K$

$1.5K=T^{o}_f-T_f=273K-T_f$

$T_f=273 K-1.5K=271.5K$

The freezing point of the solution is 271.5K