Verified Solution • 2025–26 Session moderate

QJEEM22Ch3074: JEE Main 2022 Chemistry - Class 12 JEEM22Ch3074 Solution: The molar conductivity of a conductivity cell filled with 10…

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is Λm1{\Lambda _{m1}} and that of 20 moles another identical cell heaving 80 mL NaCl solution is Λm2{\Lambda _{m2}} . The conductivities exhibited by these two cells are same. The relationship between Λm2{\Lambda _{m2}} and Λm1{\Lambda _{m1}} is

Multiple Choice Options

  • A Λm2{\Lambda _{m2}} = 2 Λm1{\Lambda _{m1}}
  • B Λm2{\Lambda _{m2}} = Λm1{\Lambda _{m1}} / 2
  • C Λm2{\Lambda _{m2}} = Λm1{\Lambda _{m1}}
  • D Λm2{\Lambda _{m2}} = 4 Λm1{\Lambda _{m1}}

Step-by-Step Verified Solution

Λm1=k1×1000M1=k×1000100.02\Lambda_{\mathrm{m}_{1}}=\frac{\mathrm{k}_{1} \times 1000}{\mathrm{M}_{1}}=\frac{\mathrm{k} \times 1000}{\frac{10}{0.02}}

Λm2=k2×1000200.08\Lambda_{\mathrm{m}_{2}}=\frac{\mathrm{k}_{2} \times 1000}{\frac{20}{0.08}}

It is given that k1=k2\mathrm{k}_{1}=\mathrm{k}_{2}

k1=Λm12k2=Λm24\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.

Λm12=Λm24Λm2=2Λm1\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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