Question

The difference in height of the mercury column in two arms of U tube manometer in the arrangement - I is $h_1=660$ mm. In another arrangement - II at same temperature, 222 gm of $CaC{l}_2$ is dissolved in 324 gm of water and difference in height of mercury column in two arms is found to be $h_2=680$ mm. If the value of degree of dissociation for $CaC{l}_2$ in arrangement - II is $\alpha$, then the value of 64 $\alpha$ is:

[Atmospheric pressure $=$ 1 atm]

• A. $40$
• B. $38$
• C. $45$
• D. $50$

Answer 1

The correct option is A $40$

$M_{CaC{l}_2}=111g$
$n_{CaC{l}_2}=\frac{222}{111}=2mole$; $n_{H_{2}O}=\frac{324}{18}=18mole$

$P_0=760-660=100$ mm of Hg

$P_s=760-680=80$ mm of Hg

Relative lowering in vapour pressure $=\frac{P_0-P_S}{P_0}=\frac{n_{1}i}{n_{1}i+n_2}=\frac{100-80}{100}=\frac{i\times 2}{i\times 2+18}$
$=>0.2=\frac{2i}{2i+18}$

$=>0.4i+3.6=2i$

$\therefore i=2.25$

For $CaC{l}_2\to i=1+(n-1)\alpha$
$2.25=1+(3-1)\alpha$
$\alpha =\frac{1.25}{2}=\mathrm{0.625.}$
$64\times 0.625$

$=40$