Question

# Calculate the freezing point of a solution containing $$60\ g$$ of glucose (Molar mass $$=180\ g\ mol^{-1}$$) in $$250\ g$$ of water. ($$Kf$$ of water $$=1.86\ K\ kg\ mol^{-1}$$).

Solution

## The number of moles of glucose $$= \dfrac { \text {Mass of glucose } }{ \text { Molecular weight of glucose} }= \dfrac { \text { 60 g } }{ \text { 180 g/mol } }=\text {0.333 mol}$$The molality of glucose solution $$= \dfrac { \text {Number of moles of glucose } }{ \text { Mass of solvent (in kg)} }= \dfrac {\text {0.333 mol} }{ \text { 250 g } \times \dfrac {\text{1 kg}}{\text{1000 g}} }=\text {1.333 mol/kg}$$The depression in the freezing point of glucose solution :$$\Delta T_f = K_f m = \text {1.86K kg /mol} \times \text {1.333 mol/kg}=\text {2.48 K}$$The freezing point of pure water is $$0^oC$$The freezing point of glucose solution $$=0-2.48=- {2.48 } ^oC$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More