Given that molar conductances for \(Ba(OH)_2\), \(BaCl_2\) and \(NH_4Cl\) are 523.28, 280.0 and 129.8 \( \Omega^{-1}cm^2mol^{-1} \), respectively. What is the molar conductivity \( \Lambda \) of \(NH_4OH\) at this temperature will be:
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- 502.88
- 251.44
- 1005.76
- 209.42
The Correct Option is B
Solution and Explanation
Molar conductivity \( \Lambda \) is related to the molar conductance of the ions in solution. To calculate the molar conductivity of \(NH_4OH\), we use the fact that: \[ \Lambda_{NH_4OH} = \Lambda_{Ba(OH)_2} + \Lambda_{NH_4Cl} - \Lambda_{BaCl_2} \] Substituting the values: \[ \Lambda_{NH_4OH} = 523.28 + 129.8 - 280.0 = 251.44 \] Thus, the correct answer is 251.44.
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