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Question

A $$10$$% solution (by mass) of sucrose in water has a freezing point of $$269.15\ K$$. Calculate the freezing point of $$10$$% glucose in water if the freezing point of pure water is $$273.15\ K$$.
(Given : molar mass of sucrose $$= 342\ g\ mol^{-}$$, Molar mass of glucose $$= 180\ g\ mol^{-1}$$)


A
275. 53 K
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B
265. 53 K
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C
271. 32 K
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D
282.43 K
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Solution

The correct option is B 265. 53 K
For sucrose,
$$\Delta T_f=K_fm$$

$$\Delta T_f=K_f\times \cfrac {10}{342}\times \cfrac {1}{Mass\quad of\quad solvent} \longrightarrow (1)$$

For glucose,
$$\Delta T_f=K_f\times \cfrac {10}{180}\times \cfrac {1}{Mass\quad of\quad solvent}\longrightarrow (2)$$

Both have the same mass of solvent

Divide equation (1) & (2)

$$\Rightarrow \cfrac {\Delta T_f (sucrose)}{\Delta T_f (glucose)}=\cfrac {180}{342}$$

$$\Rightarrow \cfrac {T_f^o-T_f (sucrose)}{T_f^o-T_f (glucose)}=\cfrac {180}{342}$$

$$\Rightarrow \cfrac {273.15-269.15}{273.15-T_f}=\cfrac {180}{342}$$

$$\Rightarrow \cfrac {4}{273.15-T_f}=0.526$$

$$\Rightarrow \cfrac {4}{0.526}=273.15- T_f$$

$$\Rightarrow T_f=273.15-\cfrac {4}{0.526}$$

$$\Rightarrow T_f= 273.15-7.6$$

$$\Rightarrow T_f=265.54K$$

Chemistry

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