JEE MainChemistryModeratePhysical Chemistry

MX2 dissociates into M2+ and X- ions in an aqueous
solution, with a degree of dissociation $(\alpha)$ of 0.5 The ratio of
the observed depres sion of freezing point of the aqueous solution to
the value of the de pression of freezing point in the absence of ionic
diss ociation is
MX2,d 0.5 dh,d fo;kstu ek=k (degree of
dissociation) α ds lkFk M2+ rFkk X- gksrk gSA
fgekad voueu (depression of freezing point)
rFkk (depression)

Show answer and solution

{MX}_{2}\ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ M^{2 +} + 2X^{-}

$m_{0}\left( \begin{matrix}
1 - \alpha)\ \ \ \ m_{0}\alpha\ \ \ \ 2m_{0}\alpha
\end{matrix} \right.\

;m = m_{0}(1 + 2\alpha)$

\therefore m = m_{0}(1 + 2 \times 0.5) = 2m_{0}

(as given) k

\frac{\left( - \Delta T_{f} \right)_{observed\ }}{\left( - \Delta T_{f} \right)_{undissociated\ }} = i = \frac{m}{m_{0}} = 2

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

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