JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 17)
If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ____________ molal. (Rounded off to the nearest integer)
[Kb = 0.52 K kg mol$$-$$1]
[Kb = 0.52 K kg mol$$-$$1]
Answer
3
Explanation
As AB is a binary electrolyte,
$$\therefore$$ AB $$\rightleftharpoons$$ A+ + B$$-$$, n = 2
$$i = 1 + \alpha (n - 1) = 1 + {{75} \over {100}}(2 - 1) = 1.75$$
Given, $$\Delta$$Tb = 2.5 K
Kb = 0.52 K kg mol$$-$$1
$$\therefore$$ $$\Delta$$Tb = Kb $$\times$$ m $$\times$$ i
$$ \Rightarrow m = {{\Delta {T_b}} \over {{K_b} \times i}} = {{2.5} \over {0.52 \times 1.75}}$$
$$ = 2.74 \simeq 3$$ mol/kg
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