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]
Answer (integer)
50
Solution
$\Delta$T<sub>f</sub> = K<sub>f</sub> (im)<br><br>$\Rightarrow$ 3.885 = i $\times$ 1.85 $\times$ 2<br><br>$\Rightarrow$ i = 1.05<br><br>Also, we know,<br><br>i = 1 + (n $-$ 1) $\alpha$<br><br>here n = number of particle obtained upon the dissociation of one particle.<br><br>$HA\rightleftharpoons H^{+}+A^{-}$<br><br>here from one particle HA we get two particle H<sup>+</sup> and A<sup>$-$</sup>.<br><br>$\therefore$ n = 2<br><br>So, i = 1 + (2 $-$ 1)$\alpha$<br><br>$\Rightarrow$ 1.05 = 1 + $\alpha$<br><br>$\Rightarrow$ $\alpha$ = 0.05 = 50 $\times$ 10<sup>$-$3</sup>
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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