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  • The molar conductivity of a conductivity cell filled with moles of <img alt="20 \mathrm{~mL}\: \mathrm{NaCl}" src="/latex-image/?20%20%5Cmathrm%7B%7EmL%7D%5C

The molar conductivity of a conductivity cell filled with 10 moles of 20 \mathrm{~mL}\: \mathrm{NaCl} solution is \Lambda_{\mathrm{m} 1} and that of 20 moles another identical cell heaving \mathrm{ 80 \mathrm{~mL}\; \;\! \! \mathrm{NaCl}} solution is \mathrm{ \Lambda_{\mathrm{m} 2},} The conductivities exhibited by these two cells are same. The relationship between \Lambda_{\mathrm{m} 2} and \mathrm{\Lambda_{\mathrm{m} 1}} is

Option: 1

\Lambda_{\mathrm{m} 2}=2 \Lambda_{\mathrm{m} 1}


Option: 2

\Lambda_{\mathrm{m} 2}=\Lambda_{\mathrm{m} 1} / 2


Option: 3

\Lambda_{\mathrm{m} 2}=\Lambda_{\mathrm{m} 1}


Option: 4

\Lambda_{\mathrm{m} 2}=4 \Lambda_{\mathrm{m} 1}


Answers (1)

best_answer

We know,

\mathrm{\Lambda_m=K \times \frac{1000}{M}}                        \begin{Bmatrix} \mathrm{M=molarity}\\ \mathrm{K=conductivity}\end{Bmatrix}

If both cell are identical, K will be same.

So,

\mathrm{\Lambda_m \times \frac{1000}{M}}

Then,

\mathrm{\frac{\Lambda_{m_1}}{\Lambda_{m_2}}=\frac{M_2}{M_1}=\frac{20}{80} \times \frac{20}{10}}

\mathrm{\frac{\Lambda_{m_1}}{\Lambda_{m_2}}=\frac{1}{2}}

\mathrm{\Lambda_{m_2}=2 \Lambda_{m_1}}

Hence, the correct answer is Option (1).

Posted by

Shailly goel

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