The freezing point depression of 0.001 m, $K_{x}\left\lbrack Fe(CN)_{6} ight brack$is $7.10 imes 10^{- 3}K$. If for water, $k_{f}$ is 1.86 K Kg $mol^{- 1}$, value of $x$ will be

Question

The freezing point depression of 0.001 m, $K_{x}\left\lbrack Fe(CN)_{6} ightbrack$is $7.10 	imes 10^{- 3}K$. If for water, $k_{f}$ is 1.86 K Kg $mol^{- 1}$, value of $x$ will be

Dr. Abubacker · Accepted Answer

(b)
$\mathrm{\Delta}x = i 	imes k_{f} 	imes m$
$7.10 	imes 10^{- 3} = i 	imes 1.86 	imes 0.001$
$i = 3.817$
$\because\ \alpha = \frac{i - 1}{n - 1}$
$	herefore 1 = \frac{3.817 - 1}{(x + 1) - 1}$
$x = 2.817 \approx 3$
$	herefore$ molecular formula of the compound is $K_{3}\lbrack Fe(CN)_{6}brack$

Q. The freezing point depression of 0.001 m, $K_{x}\left\lbrack Fe(CN)_{6} \right\rbrack$is $7.10 \times 10^{- 3}K$. If for water, $k_{f}$ is 1.86 K Kg $mol^{- 1}$, value of $x$ will be

Explanation

Reviewed by Dr. Abubacker

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