Q36 / 225 · JEE Main 2004 · Chemistry

The limiting molar conductivities $\Lambda^{\circ}$ for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126,152 and $150 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The $\Lambda^{\circ}$ for $\mathrm{NaBr}$ is

  • Correct answer
    A
    $128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • B
    $302 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • C
    $278 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • D
    $176 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

$\Lambda_{\mathrm{NaCl}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Cl}}^{\circ}=126 \ldots(1)$ $\Lambda_{\mathrm{KBr}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}=152$ $\Lambda_{\mathrm{KCl}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}=150$ $\Lambda_{\mathrm{NaBr}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}$ $\Lambda_{\mathrm{NaBr}}^{\circ}=126+152-150=128$

Hence, the correct answer is option: 'A'.

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