JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 20)
AB2 is 10% dissociated in water to A2+ and B$$-$$. The boiling point of a 10.0 molal aqueous solution of AB2 is __________$$^\circ$$C. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water Kb = 0.5 K kg mol$$-$$1 boiling point of pure water = 100$$^\circ$$C]
[Given : Molal elevation constant of water Kb = 0.5 K kg mol$$-$$1 boiling point of pure water = 100$$^\circ$$C]
Answer
106
Explanation
AB2 $$ \leftrightharpoons $$ A+ + 2B$$-$$
$$ \therefore $$ For AB2, n = 3
i = 1 + (n $$-$$ 1)$$\alpha$$
= 1 + (3 $$-$$ 1) $$\times$$ 0.1
= 1.2
Now, $$\Delta$$Tb = Kb (im)
$$ \Rightarrow $$ Tb $$-$$ T$$_b^o$$ = 1.2 $$\times$$ 0.5 $$\times$$ 10
$$ \Rightarrow $$ Tb $$-$$ 100 = 6
$$ \Rightarrow $$ Tb = 106
$$ \therefore $$ For AB2, n = 3
i = 1 + (n $$-$$ 1)$$\alpha$$
= 1 + (3 $$-$$ 1) $$\times$$ 0.1
= 1.2
Now, $$\Delta$$Tb = Kb (im)
$$ \Rightarrow $$ Tb $$-$$ T$$_b^o$$ = 1.2 $$\times$$ 0.5 $$\times$$ 10
$$ \Rightarrow $$ Tb $$-$$ 100 = 6
$$ \Rightarrow $$ Tb = 106
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