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  4. 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to -4° C before freezing. The amount of ice (in g) that will be separated out is (Nearest integer) [. Given: .K f ( H 2 O )=1.86 K kg mol -1]

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Q. $1\, kg$ 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)
$\left[\right.$ Given : $\left.K _{ f }\left( H _{2} O \right)=1.86 \,K \,kg \,mol ^{-1}\right]$

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Solution:

Let mass of water initially present $= x gm$
$\Rightarrow $ Mass of sucrose $=(1000- x ) gm$
$\Rightarrow $ moles of sucrose $=\left(\frac{1000- x }{342}\right)$
$\Rightarrow 0.75=\frac{\left(\frac{1000- x }{342}\right)}{\left(\frac{ x }{1000}\right)} $
$\Rightarrow \frac{ x }{1000}=\frac{1000- x }{342 \times 0.75}$
$\Rightarrow 256.5 x =10^{6}-1000 x$
$\Rightarrow x =795.86\, gm$
$\Rightarrow $ moles of sucrose $=0.5969$
New mass of $H _{2} O = a \,kg$
$\Rightarrow 4=\frac{0.5969}{a} \times 1.86 $
$\Rightarrow a=0.2775\,kg$
$\Rightarrow $ ice separated $=(795.86-277.5)$
$=518.3 \,gm$