JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 4)
The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is $${\Lambda _{m1}}$$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is $${\Lambda _{m2}}$$. The conductivities exhibited by these two cells are same. The relationship between $${\Lambda _{m2}}$$ and $${\Lambda _{m1}}$$ is
$${\Lambda _{m2}}$$ = 2$${\Lambda _{m1}}$$
$${\Lambda _{m2}}$$ = $${\Lambda _{m1}}$$ / 2
$${\Lambda _{m2}}$$ = $${\Lambda _{m1}}$$
$${\Lambda _{m2}}$$ = 4$${\Lambda _{m1}}$$
Explanation
$$\Lambda_{\mathrm{m}_{1}}=\frac{\mathrm{k}_{1} \times 1000}{\mathrm{M}_{1}}=\frac{\mathrm{k} \times 1000}{\frac{10}{0.02}}$$
$$ \Lambda_{\mathrm{m}_{2}}=\frac{\mathrm{k}_{2} \times 1000}{\frac{20}{0.08}} $$
It is given that $$\mathrm{k}_{1}=\mathrm{k}_{2}$$
$$ \mathrm{k}_{1}=\frac{\Lambda_{\mathrm{m}_{1}}}{2} \quad \quad \mathrm{k}_{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4} $$
Applying the given condition on conductivity.
$$ \begin{gathered} \frac{\Lambda_{\mathrm{m}_{1}}}{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4} \\ \Lambda_{\mathrm{m}_{2}}=2 \Lambda_{\mathrm{m}_{1}} \end{gathered} $$
$$ \Lambda_{\mathrm{m}_{2}}=\frac{\mathrm{k}_{2} \times 1000}{\frac{20}{0.08}} $$
It is given that $$\mathrm{k}_{1}=\mathrm{k}_{2}$$
$$ \mathrm{k}_{1}=\frac{\Lambda_{\mathrm{m}_{1}}}{2} \quad \quad \mathrm{k}_{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4} $$
Applying the given condition on conductivity.
$$ \begin{gathered} \frac{\Lambda_{\mathrm{m}_{1}}}{2}=\frac{\Lambda_{\mathrm{m}_{2}}}{4} \\ \Lambda_{\mathrm{m}_{2}}=2 \Lambda_{\mathrm{m}_{1}} \end{gathered} $$
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