Question

The depression of freezing point of a 3 % aqueous solution of a substance X is equal to the depression of freezing point of a 2 % aqueous solution of a substance Y. If the molecular weight of Y is 60, then the molecular weight of X is *

60.23
90.92
40.54
120.12

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Answer 1

Answer:

The correct option is 40.54

Explanation:

Hope it helps you.......

Answer 2

Answer:

x = 90.92

Option B ) is correct

Explanation:

Δ$T_{f}$ = $Kf \frac{W_{B} }{M_{B} W_{A} } *100$

Let molecular weight of x = a

3 % aqueous solution means 3 g of solute in 100g solution

Then Δ$T_{f}$ (in X) = $\frac{3Kf}{x*97} *1000$

2 % aqueous solution means 2 g of solute in 100g of solution

$W_{A}$=100-2 = 98 = 98*10⁻³kg

Δ$T_{f}$ (in Y) = $\frac{2Kr}{60*98} *1000$

since ΔT is equal for two solutions,

$\frac{3Kf}{x*97} *1000$  =   $\frac{2Kr}{60*98} *1000$

$x = \frac{3 *60 *98}{2*97}$

$x = \frac{17640}{194}$

x = 90.92

Option B ) is correct