The conductivity of 0.1 M NaOH is 0.022 S cm−1. When equal volume of 0.1 M HCl is added to this solution, the conduction of the solution becomes 0.00632 S cm−1. The molar conductivity of sodium chloride in the resultant solution is

# The conductivity of 0.1 M NaOH is . When equal volume of 0.1 M HCl is added to this solution, the conduction of the solution becomes . The molar conductivity of sodium chloride in the resultant solution is

1. A

2. B

3. C

4. D

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### Solution:

The concentration of NaCl in the resultant solution will be 0.05 M. Hence,

$\mathrm{\Lambda }=\frac{\kappa }{c}=\frac{0.00632\mathrm{S}{\mathrm{cm}}^{-1}}{0.05\mathrm{mol}{\mathrm{dm}}^{-3}}=0.1264\mathrm{S}{\mathrm{cm}}^{-1}{\mathrm{mol}}^{-1}{\mathrm{dm}}^{3}=0.1264\mathrm{S}{\mathrm{cm}}^{-1}{\mathrm{mol}}^{-1}\left(10\mathrm{cm}{\right)}^{3}=126.4\mathrm{S}{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}$

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