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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

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