Question

The molar conductivity of a 0.5 mol/dm3 solution of agno3

Answer 1

Answer:

$lamda \: m \: = k \times 1000 \div molarity \\ = 5.76 \times 10 { - 3 \: } times 1000 \div 0.5 = 11.52$

Answer 2

The given question is incomplete. The complete question is as follows.

The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76 x 10-3 S cm-1 at 298 K is

The molar conductivity of given solution is 11.52 $cm^{2}/mol$.

Explanation:

It is given that molarity is 0.5 $mol/dm^{3}$ and value of electrolytic conductivity is $5.76 \times 10^{-3} S cm^{-1}$ at 298 K.

Therefore, relation between molar conductivity and molarity is as follows.

            $\lambda^{o} = \frac{\kappa \times 1000}{\text{molarity}}$

Hence, putting the given values into the above formula as follows.

          $\lambda^{o} = \frac{\kappa \times 1000}{\text{molarity}}$

                      = $\frac{5.76 \times 10^{-3} S cm^{-1} \times 1000}{0.5 mol/dm^{3}}$

                      = 11.52 $cm^{2}/mol$

Thus, we can conclude that the molar conductivity of given solution is 11.52 $cm^{2}/mol$.

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