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← JEE Main 2022 · 29 Jul · Shift 1
ChemistryInteger+4 / -1Question 81 of 90

JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 81

Resistance of a conductivity cell (cell constant 129 m−1^{-1}) filled with 74.5 ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. Λ1Λ2=x×10−3\dfrac{\Lambda_1}{\Lambda_2}=x\times10^{-3}. The value of xx is _______. (Nearest integer)

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

Answer and worked solution

Official answer: 1000

Concentrations. 74.5 ppm is 74.5 mg/L, i.e. 1×10−31\times10^{-3} M; 149 ppm is 2×10−32\times10^{-3} M.

Conductivities 129100=1.29\frac{129}{100}=1.29 and 12950=2.58\frac{129}{50}=2.58 S/m. Then Λ1=1.2910−3\Lambda_1=\frac{1.29}{10^{-3}}, Λ2=2.582×10−3\Lambda_2=\frac{2.58}{2\times10^{-3}}: equal, ratio 1=1000×10−31=1000\times10^{-3}.

Answer: 1000

Topics:Conductance and Kohlrausch's lawConcentration terms
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