Question

100 ml of 0.1 M solution of a weak acid $HA$ has molar conductivity $5Sc{m}^{2}mo{l}^{-1}$. The osmotic pressure of the resulting solution obtained after dilution of original solution upto 1 litre at 500 K, assuming ideal solution is:
$(Given:{\lambda }_m^{\infty }(H^{+})=450Sc{m}^{2}mo{l}^{-1},{\lambda }_m^{\infty }(A^{-})=50Sc{m}^{2}mo{l}^{-1}$ , $R=0.08Latmmol{e}^{-1}{K}^{-1}$, $\sqrt{10}=3.2)$

• A. 0.3128 atm
• B. 0.4128 atm
• C. 0.5128 atm
• D. 0.6128 atm

Answer 1

The correct option is B 0.4128 atm

For original solution, $\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\infty }}=\frac{5}{450+50}=\frac{1}{100}.$
For dilution, $C_1{V}_1=C_2{V}_2$ $\Rightarrow 100\times 0.1=1000\times C_2.$
$\Rightarrow C_2=0.01M.$
For weak acid,

$HA\rightleftharpoons H^{+}+A^{-}$
$C$

$C(1-\alpha )C\alpha C\alpha$

Now,
$K_a=C_1{\alpha }_1^2=C_2{\alpha }_2^2$
$0.1\times (\frac{1}{100}{)}^2=0.01{\alpha }_2^2$
${\alpha }_2=0.032$
Again,
Osmotic pressure $=i\times cRT=(1+{\alpha }_2)\times 0.01\times 0.08\times 500$$=0.4128$ atm.