JEE MAIN - Chemistry (2006 - No. 33)
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100$$\Omega $$.
The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of
the same solution is 520 $$\Omega $$, The molar conductivity of 0.02 M solution of the electrolyte will be
124 $$\times$$ 10–4 S m2 mol–1
1240 $$\times$$ 10–4 S m2 mol–1
1.24 $$\times$$ 10–4 S m2 mol–1
12.4 $$\times$$ 10–4 S m2 mol–1
Explanation
$$R = 100\Omega ,\kappa = {1 \over R}\left( {{l \over a}} \right),{l \over a}$$ (cell constant)
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 1.29 \times 100{m^{ - 1}}$$
Given, $$R = 520\Omega ,C = 0.2M,\,\mu $$ (molar conductivity) $$=$$ ?
$$\mu = \kappa \times V\,\,\,\,\,\,\,\,$$ ($$\kappa $$ can be calculated as $$\kappa = {1 \over R}\left( {{1 \over a}} \right)$$
now cell constant is known.
Hence, $$\mu = {1 \over {520}} \times 129 \times {{1000} \over {0.2}} \times {10^{ - 6}}{m^3}$$
$$ = 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 1.29 \times 100{m^{ - 1}}$$
Given, $$R = 520\Omega ,C = 0.2M,\,\mu $$ (molar conductivity) $$=$$ ?
$$\mu = \kappa \times V\,\,\,\,\,\,\,\,$$ ($$\kappa $$ can be calculated as $$\kappa = {1 \over R}\left( {{1 \over a}} \right)$$
now cell constant is known.
Hence, $$\mu = {1 \over {520}} \times 129 \times {{1000} \over {0.2}} \times {10^{ - 6}}{m^3}$$
$$ = 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}}$$
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