Question

100 ml of 0.1 M solution of a weak acid HA has molar conductivity 5 $Sc{m}^{2}mo{l}^{-1}$. Calculate the osmotic pressure of resulting solution obtained after dilution of original solution upto 1 litre at 500 K assuming ideal solution.

Report your answer as X, where X $=$ Osmotic pressure (in atm) $\times$ 10 (nearest integer).

$(Given:{\lambda }_m^{\infty }(H^{+})=450Sc{m}^{2}mo{l}^{-1},{\lambda }_m^{\infty }(A^{-})=50Sc{m}^{2}mo{l}^{-1},R=0.08Latm/molK,\sqrt{10}=3.2).$

Answer 1

The molar conductivity at infinite dilution is ${\lambda }_{\infty }=450Sc{m}^2/mol+50Sc{m}^2/mol=500Sc{m}^2/mol$

The degree of dissociation is $\alpha =\frac{\lambda }{{\lambda }_{\infty }}=\frac{5Sc{m}^2/mol}{500Sc{m}^2/mol}=0.01$

Hence, the Van't Hoff factor $i=1.01$.

100 mL of the solution is diluted to 1 L. The dilution is 10 times. Hence, the concentration will be one-tenth.

It will be $\frac{0.1M}{10}=0.01M$.

The osmotic pressure of the solution is:

$\Pi =CRT=1.01\times 0.01M\times 0.08L\cdot atm/mol/K\times 500K=0.404atm$

Hence, $X=10\times 0.4=4$.