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← JEE Main 2026 · 5 Apr · Shift 2
ChemistryInteger+4 / -1Question 74 of 75

JEE Main 2026 · 5 Apr · Shift 2, Chemistry question 74

At 298 K, the molar conductivity of xx% (w/w) MX solution (aqueous) is 123.5 S cm² mol−1^{-1}. The conductance of same solution is 1.9×10−31.9\times10^{-3} S. The value of xx is ________ ×10−2\times10^{-2}.

(Given : cell constant = 1.3 cm−1^{-1}; molar mass of MX is 75 g mol−1^{-1}, density of aqueous solution of MX at 298 K is 1.0 g mL−1^{-1})

Answer and worked solution

Official answer: 15

Conductivity. κ=G×G∗=1.9×10−3×1.3=2.47×10−3\kappa=G\times G^*=1.9\times10^{-3}\times1.3=2.47\times10^{-3} S cm−1^{-1}.

Molarity.
c=1000κΛm=2.47123.5=0.02 M.c=\frac{1000\kappa}{\Lambda_m}=\frac{2.47}{123.5}=0.02\text{ M}.

Mass percent. 1 L of solution weighs 1000 g and holds 0.02×75=1.50.02\times75=1.5 g of MX. That is 0.15%, or 15×10−215\times10^{-2}.

Answer: 15

Topics:Conductance and Kohlrausch's lawConcentration terms
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