Question

For $HCl$ solution, at $25{}^{\circ }\text{C}$ the equivalent conductivity at infinte dilution is $425{\Omega }^{-1}{\text{cm}}^{-1}$. The specific conductance of a solution of $HCl$ is $3.825{\Omega }^{-1}{\text{cm}}^{-1}$. If the apparent degree of dissosiation is $90\%$. The normality of solution is:

• A. $9\text{N}$
• B. $10\text{N}$
• C. $11\text{N}$
• D. $12\text{N}$

Answer 1

The correct option is B $10\text{N}$

Given, equivalent conductivity at infinite dilution, ${\wedge }_{eq}^{\propto }=425{\Omega }^{-1}{\text{cm}}^2{\text{eq}}^{-1}$
Degree of dissociation $\left(\alpha \right)=90\text{\%}or0.9$
Specific conductance of solution ($\kappa$)
$=3.825{\Omega }^{-1}{\text{cm}}^{-1}$
$\therefore \alpha =\frac{{\wedge }_{eq}}{{\wedge }_{eq}^{\propto }}$
$({\wedge }_{eq}=$ equivalent conductance at the particular concentration C)
$\therefore {\wedge }_{eq}$ at concentration $C$ $=0.9\times 425=382.5{\Omega }^{-1}{\text{cm}}^2{\text{eq}}^{-1}$
Now,$C=\frac{\kappa \times 1000}{{\wedge }_{eq}}=\frac{3.825}{382.5}\times 1000=10\text{g equivalent/L}$
Thus normality of the solution is $10\text{N}$