Question

$200ml$ of $0.2M$ solution of a weak acid $HA$ has molar conductivity $10Sc{m}^{2}mo{l}^{-1}$. The osmotic pressure of the resulting solution obtained after dilution of original solution upto 1 litre at $500K$, 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. $0.4168atm$
• B. $0.3128atm$
• C. $1.6720atm$
• D. $0.5128atm$

Answer 1

Answer: (C)

$1.6720atm$

For original solution, $\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\infty }}=\frac{10}{450+50}=\frac{1}{50}.$

For dilution, $C_1{V}_1=C_2{V}_2$ $\Rightarrow 200\times 0.2=1000\times C_2.$

$\Rightarrow C_2=0.04M.$

For weak acid, $HA\rightleftharpoons H^{+}+A^{-}$

$C$ 0 0

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

Now, $K_a=C_1{\alpha }_1^2=C_2{\alpha }_2^2$

$0.2\times (\frac{1}{50}{)}^2=0.02{\alpha }_2^2$

${\alpha }_2=0.045$

Again, Osmotic pressure $=i\times cRT=(1+{\alpha }_2)\times 0.04\times 0.08\times 500$$=1.672$atm.

Option C is correct.