KCET · Chemistry · Electrochemistry
Specific conductance of \(0.1 \mathrm{M} \mathrm{HNO}_3\) is \(63 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The molar conductance of the solution is
- A \(315 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- B \(6.300 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- C \(63.0 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
- D \(630 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(630 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, specific conductance,
\[
\begin{aligned}
k=6.3 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \\
\mathrm{HNO}_3 \text { concentration, } c=0.1 \mathrm{M} \\
\text { Molar conductance } \Lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathcal{c}} \\
=6.3 \times 10^{-2} \times \frac{1000}{0 .}=630 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
\]
\[
\begin{aligned}
k=6.3 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \\
\mathrm{HNO}_3 \text { concentration, } c=0.1 \mathrm{M} \\
\text { Molar conductance } \Lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathcal{c}} \\
=6.3 \times 10^{-2} \times \frac{1000}{0 .}=630 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
\]
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