JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 13)
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -
4A
2A
3A
A
Explanation
For same freezing point,
($$\Delta $$Tf)X = ($$\Delta $$Tf)Y
$$ \Rightarrow $$ kf mx = kf my
$$ \Rightarrow $$ $${{4 \times 1000} \over {A \times 96}} = {{12 \times 1000} \over {M \times 88}}$$
$$ \Rightarrow $$ M = 3.27A $$ \simeq $$ 3A
($$\Delta $$Tf)X = ($$\Delta $$Tf)Y
$$ \Rightarrow $$ kf mx = kf my
$$ \Rightarrow $$ $${{4 \times 1000} \over {A \times 96}} = {{12 \times 1000} \over {M \times 88}}$$
$$ \Rightarrow $$ M = 3.27A $$ \simeq $$ 3A
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