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Chemistry · Electrochemistry

JEE Main 2026 — 5 April, Evening Shift — Question 69

At 298 K , the molar conductivity of x%(w/w)\mathrm{x} \%(\mathrm{w} / \mathrm{w}) MX solution (aqueous) is 123.5 S cm2 mol−1123.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}. The conductance of same solution is 1.9×10−3 S1.9 \times 10^{-3} \mathrm{~S}. The value of xx is ____\_\_\_\_ ×10−2\times 10^{-2}. (Given : cell constant =1.3 cm−1=1.3 \mathrm{~cm}^{-1}; molar mass of MX is 75 g mol−175 \mathrm{~g} \mathrm{~mol}^{-1}, density of aqueous solution of MX at 298 K is 1.0 g mL−11.0 \mathrm{~g} \mathrm{~mL}^{-1} )

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

Let molarity of MX solution =yM=\mathrm{y} \mathrm{M} ⇒Λm=c(ℓ/A)×1000y\Rightarrow \Lambda_{\mathrm{m}}=\mathrm{c}(\ell / \mathrm{A}) \times \frac{1000}{\mathrm{y}} ⇒123.5=1.9×10−3×1.3×1000y\Rightarrow 123.5=1.9 \times 10^{-3} \times 1.3 \times \frac{1000}{\mathrm{y}} ⇒y=0.02M\Rightarrow \mathrm{y}=0.02 \mathrm{M} Let molar mass of MX=Mgmol−1\mathrm{MX}=\mathrm{M} \mathrm{g} \mathrm{mol}^{-1} 0.2 mol MX is dissolve in 1 L solution ( 0.02×750.02 \times 75 ) gm MX is dissolved in 1000 g solution %w/w=x%=0.02×751000×100\% \mathrm{w} / \mathrm{w}=\mathrm{x} \%=\frac{0.02 \times 75}{1000} \times 100 =0.15%=0.15 \%

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Conductance of Solutions and Kohlrausch's Law

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