The difference in height of the mercury column in two arms of U tube manometer in arrangement-I is $h_{1} = 660 m m$ . In another arrangement-II at same temperature, 222 gm of $C a C l_{2}$ , is dissolved in 324 gm of water and difference in height of mercury column in two arms is found to be $h_{2} = 680 m m$ . If the value of degree of dissociation for $C a C l_{2}$ , in arrangement-II is $α$ then the value of $6.4 α$ is :[Atmospheric pressure = 1atm]

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Detailed Solution

$M_{C a C l_{2}} = 111 g$
$n_{C a C l_{2}} = \frac{222}{111} = 2 m o l e$ $n_{H_{2} O} = \frac{324}{18} = 18 m o l e$
Relative lowering in vapour pressure
$R . L . V . P = \frac{P_{0} - P_{S}}{P_{0}} = \frac{n_{1} i}{n_{1} i + n_{2}} = \frac{100 - 80}{100} = \frac{i \times 2}{i \times 2 + 18}$
or $0.2 = \frac{2 i}{2 i + 18}$ or $0.4 i + 3.6 = 2 i$
So $i = 2.25$
For $C a C l_{2} \overset{}{\to} i = 1 + (n - 1) α$
$2.25 = 1 + (3 - 1) α$
$α = \frac{1.25}{2} = 0.625$
$6.4 \times 0.625 = 4$