The freezing point of aqueous solution that contains $3$ % urea, $7.45$ % $K C l$ and $9$ % of glucose is: (given $K_{1}$ of water $= 1.86$ and assume $m o l a r i t y = m o l a l i t y$ ).

290 K

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285.5 K

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267.42 K

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250 K

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Solution

The correct option is B $267.42 K$
urea = $3 %$
KCl = $7.45 %$
Glucose = $9 %$
Let Amount of water be $100 g m N H_{2} C O N H_{2}$ urea = $3 g m$ molar mass = $60$
$K C l = 7.45 g$ molar mass = $74.5$ $C_{6} H_{12} O_{6}$ Glucose = $9$ g Molar mass = $180$
total no of mole = $\frac{3}{60} + \frac{7.45}{74.5} + \frac{9}{180}$
$\frac{1}{20} + \frac{1}{10} + \frac{1}{20}$
$= \frac{1}{5} = 0.2$
molality = $\frac{0.2}{100} = 2 m$
$Δ T_{f} = k_{f} m$
$= 1.86 \times 2 = 3.72$
Freezing point = $273 - 3.72$
$= 269.28 k$

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The freezing point of aqueous solution that contains 3% urea, 7.45% KCl and 9% of glucose is: (given K1 of water =1.86 and assume molarity=molality).

The freezing point of aqueous solution that contains $3$ % urea, $7.45$ % $K C l$ and $9$ % of glucose is: (given $K_{1}$ of water $= 1.86$ and assume $m o l a r i t y = m o l a l i t y$ ).