JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 22)
2 molal solution of a weak acid HA has a freezing point of 3.885$$^\circ$$C. The degree of dissociation of this acid is ___________ $$\times$$ 10$$-$$3. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol$$-$$1 Freezing point of pure water = 0$$^\circ$$ C]
[Given : Molal depression constant of water = 1.85 K kg mol$$-$$1 Freezing point of pure water = 0$$^\circ$$ C]
Answer
50
Explanation
$$\Delta$$Tf = Kf (im)
$$ \Rightarrow $$ 3.885 = i $$\times$$ 1.85 $$\times$$ 2
$$ \Rightarrow $$ i = 1.05
Also, we know,
i = 1 + (n $$-$$ 1) $$\alpha$$
here n = number of particle obtained upon the dissociation of one particle.
$$HA\rightleftharpoons H^{+}+A^{-} $$
here from one particle HA we get two particle H+ and A$$-$$.
$$ \therefore $$ n = 2
So, i = 1 + (2 $$-$$ 1)$$\alpha$$
$$ \Rightarrow $$ 1.05 = 1 + $$\alpha$$
$$ \Rightarrow $$ $$\alpha$$ = 0.05 = 50 $$\times$$ 10$$-$$3
$$ \Rightarrow $$ 3.885 = i $$\times$$ 1.85 $$\times$$ 2
$$ \Rightarrow $$ i = 1.05
Also, we know,
i = 1 + (n $$-$$ 1) $$\alpha$$
here n = number of particle obtained upon the dissociation of one particle.
$$HA\rightleftharpoons H^{+}+A^{-} $$
here from one particle HA we get two particle H+ and A$$-$$.
$$ \therefore $$ n = 2
So, i = 1 + (2 $$-$$ 1)$$\alpha$$
$$ \Rightarrow $$ 1.05 = 1 + $$\alpha$$
$$ \Rightarrow $$ $$\alpha$$ = 0.05 = 50 $$\times$$ 10$$-$$3
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