Verified Solution • 2025–26 Session moderate

QJEEM22Chc062: JEE Main 2022 Chemistry - Class 12 JEEM22Chc062 Solution: Resistance of a conductivity cell (cell constant ) filled with…

Resistance of a conductivity cell (cell constant 129 m−1129 \mathrm{~m}^{-1} ) filled with 74.5 ppm74.5 \,\mathrm{ppm} solution of KCl\mathrm{KCl} is 100 Ω100 \,\Omega (labelled as solution 1). When the same cell is filled with KCl\mathrm{KCl} solution of 149 ppm149 \,\mathrm{ppm} , the resistance is 50 Ω50 \,\Omega (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. ∧1∧2=x×10−3\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3} . The value of xx is __________. (Nearest integer)

Given, molar mass of KCl\mathrm{KCl} is 74.5 g mol−174.5 \mathrm{~g} \mathrm{~mol}^{-1} .

Step-by-Step Verified Solution

lA=129 m−1\frac{l}{A}=129 \mathrm{~m}^{-1}

KCl\mathrm{KCl} solution 1⇒74.5 ppm,R1=100Ω1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega

KCl\mathrm{KCl} solution 2⇒149 ppm,R2=50Ω2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega

amp; Here, ppm1ppm2=M1M2=(w1/M0V×Vw2/M0)amp;Λ1Λ2=k1×1000M1k2×1000M2amp;=k1k2×M1M2amp;=50100×2amp;=Λ1Λ2=1000×10−3amp;=1000\begin{aligned}&\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\&\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\&=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\&=\frac{50}{100} \times 2 \\&=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\&=1000\end{aligned}

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