JEE MAIN - Chemistry (2021 - 27th August Morning Shift - No. 16)
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 : Kf(H2O) = 1.86 K kg mol$$-$$1]
[Given : Kf(H2O) = 1.86 K kg mol$$-$$1]
Answer
518
Explanation
Let mass of water initially present = x gm
$$\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm
$$\Rightarrow$$ moles of sucrose = $$\left( {{{1000 - x} \over {342}}} \right)$$
$$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$$
$$\Rightarrow$$ 256.5x = 106 $$-$$ 1000x
$$\Rightarrow$$ x = 795.86 gm
$$\Rightarrow$$ moles of sucrose = 0.5969
New mass of H2O = a kg
$$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $$ a = 0.2775 kg
$$\Rightarrow$$ ice separated = (795.86 $$-$$ 277.5) = 518.3 gm
$$\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm
$$\Rightarrow$$ moles of sucrose = $$\left( {{{1000 - x} \over {342}}} \right)$$
$$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$$
$$\Rightarrow$$ 256.5x = 106 $$-$$ 1000x
$$\Rightarrow$$ x = 795.86 gm
$$\Rightarrow$$ moles of sucrose = 0.5969
New mass of H2O = a kg
$$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $$ a = 0.2775 kg
$$\Rightarrow$$ ice separated = (795.86 $$-$$ 277.5) = 518.3 gm
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