Question

100 ml of 0.1 M solution of a weak acid HA has molar conductivity 5 $Sc{m}^{2}mo{l}^{-1}$. The osmotic pressure of resulting solution obtained after dilution of original solution upto 1 litre at 500 K assuming ideal solution is
$(Given:{\lambda }_m^{\infty }(H^{+})=450Sc{m}^{2}mo{l}^{-1},{\lambda }_m^{\infty }(A^{-})=50Sc{m}^{2}mo{l}^{-1},R=0.08lt-atm/mole-K,\sqrt{10}=3.2).$
Report your answer as X, where X $=$ Osmotic pressure (in atm) $\times$ 10. Give your answer in nearest intiger.

Answer 1

For original solution, $\alpha =\frac{{\wedge }_m}{{\wedge }_m^{oo}}=\frac{5}{500}=\frac{1}{100}.$
For diultion, $C_1{V}_1=C_2{V}_2$ $\Rightarrow 100\times 0.1=1000\times C_2.$
$\Rightarrow C_2=0.01M.$
For weak acid, $HA\rightleftharpoons H^{+}+A^{-}$
$C(1-\alpha )C\alpha C\alpha$
$K_a=C_1{\alpha }_1^2=C_2{\alpha }_2^2$
$0.1\times (\frac{1}{100}{)}^2=0.01{\alpha }_2^2$
${\alpha }_2=\mathrm{0.032.}$
Osmotic pressure $=i\times cRT=(1+{\alpha }_2)\times 0.01\times 0.08\times 500.$
$=0.4128$ atm.