1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $- 4^{\circ} C$ before freezing. The amount of ice (in g ) that will be separated out is $____$ .
(Nearest integer) [Given : K_f(H₂O) = 1.86 K kg mol^-1 ]

Question

1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $- 4^{\circ} C$ before freezing. The amount of ice (in g ) that will be separated out is $____$ .

(Nearest integer) [Given : K_f(H₂O) = 1.86 K kg mol^-1 ]

TestHub · Accepted Answer

Let the mass of water initially present be $xg m$ .

This implies the mass of sucrose is $(1000 - x) g m$ .

Therefore, the moles of sucrose are $(\frac{1000 - x}{342})$ .

We have the equation:

$0.75 = \frac{( \frac{1000 - x}{342} )}{( \frac{x}{1000} )}$

This simplifies to:

$\frac{x}{1000} = \frac{1000 - x}{342 \times 0.75}$

$256.5 x = 1 0^{6} - 1000 x$

$x = 795.86 g m$

The moles of sucrose are $0.5969$ .

Let the new mass of H₂O be $ak g$ .

$4 = \frac{0.5969}{a} \times 1.86$

$a = 0.2775 k g$

The mass of ice separated is $(795.86 - 277.5) = 518.3 g m$ .

Chemistry - Liquid Solution Question with Solution | TestHub

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