Question

The specific conductivity of a 0.5 M aq. solution of monobasic acid, HA at ${27}^{\circ }C$ is $0.006Sc{m}^{-1}$. It's molar conductivity at infinite dilution is $200Sc{m}^2$. Calculate osmotic pressure (in atm ) if 0.5 N HA(aq) solution is at ${27}^{\circ }$C to the nearest integer. Given R= 0.08 L-atm/K-mol.

Answer 1

The relationship between the specific conductivity $\kappa$ and the molar conductivity $\Lambda$ is $\Lambda =\frac{1000\times \kappa }{C}=\frac{1000\times 0.006S/cm}{0.5M}=12Sc{m}^2/mol$.
The degree of dissociation is $\alpha =\frac{\Lambda }{\Lambda \infty }=\frac{12Sc{m}^2/mol}{200Sc{m}^2/mol}=0.06$.
The van't Hoff factor is $i=1+(n^{\prime }-1)\alpha =1+(2-1)0.06=1.06$.
The osmotic pressure is $\Pi =iCRT=1.06\times 0.5M\times 0.08L-atm/mol/K\times 300K=12.72atm\simeq 13atm$.