JEE MAIN - Chemistry (2014 (Offline) - No. 1)
Resistance of 0.2 M solution of an electrolyte is 50 $$\Omega$$. The specific conductance of the solution is 1.4 S m-1. The resistance of 0.5 M solution of the same electrolyte is 280 $$\Omega$$. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is :
5 × 103
5 × 102
5 × 10-4
5 × 10-3
Explanation
Given for $$0.2$$ $$M$$ solution
$$R = 50\Omega $$
$$\kappa = 1.45\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}\,S\,c{m^{ - 1}}$$
Now, $$R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}$$
$$ \Rightarrow {\ell \over a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}}$$
For $$0.5$$ $$M$$ solution
$$R = 280\,\Omega ;\,\kappa = ?$$
$${\ell \over a} = 50 \times 1.4 \times {10^{ - 2}}$$
$$ \Rightarrow R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}$$
$$ \Rightarrow \,\,\,\,\kappa = {1 \over {280}} \times 50 \times 1.4 \times {10^{ - 2}}$$
$$ = {1 \over {280}} \times 70 \times {10^{ - 2}} = 2.5 \times {10^{ - 3}}\,S\,c{m^{ - 1}}$$
Now, $${\Lambda _m} = {{\kappa \times 1000} \over M} = {{2.5 \times {{10}^{ - 3}} \times 1000} \over {0.5}}$$
$$ = 5\,S\,c{m^2}mo{l^{ - 1}} = 5 \times {10^{ - 4}}\,S\,{m^2}\,mo{l^{ - 1}}$$
$$R = 50\Omega $$
$$\kappa = 1.45\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}\,S\,c{m^{ - 1}}$$
Now, $$R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}$$
$$ \Rightarrow {\ell \over a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}}$$
For $$0.5$$ $$M$$ solution
$$R = 280\,\Omega ;\,\kappa = ?$$
$${\ell \over a} = 50 \times 1.4 \times {10^{ - 2}}$$
$$ \Rightarrow R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}$$
$$ \Rightarrow \,\,\,\,\kappa = {1 \over {280}} \times 50 \times 1.4 \times {10^{ - 2}}$$
$$ = {1 \over {280}} \times 70 \times {10^{ - 2}} = 2.5 \times {10^{ - 3}}\,S\,c{m^{ - 1}}$$
Now, $${\Lambda _m} = {{\kappa \times 1000} \over M} = {{2.5 \times {{10}^{ - 3}} \times 1000} \over {0.5}}$$
$$ = 5\,S\,c{m^2}mo{l^{ - 1}} = 5 \times {10^{ - 4}}\,S\,{m^2}\,mo{l^{ - 1}}$$
Comments (0)
