Skip to content
dronacharya.ai
← JEE Advanced 2017 · Paper 1
ChemistryInteger+3Question 27 of 54

JEE Advanced 2017 · Paper 1, Chemistry question 9

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm21\ \mathrm{cm^2}. The conductance of this solution was found to be 5×10−75 \times 10^{-7} S. The pH of the solution is 4. The value of limiting molar conductivity (Λmo)(\Lambda_m^o) of this weak monobasic acid in aqueous solution is Z×102 S cm−1 mol−1Z \times 10^2\ \mathrm{S\,cm^{-1}\,mol^{-1}}. The value of ZZ is

Answer and worked solution

Official answer: 6

Conductivity.
κ=GlA=5×10−7×120=6×10−5 S cm−1.\kappa=G\frac lA=5\times10^{-7}\times120=6\times10^{-5}\ \mathrm{S\,cm^{-1}}.

Molar conductivity.
Λm=1000κc=1000×6×10−50.0015=40 S cm2 mol−1.\Lambda_m=\frac{1000\kappa}c=\frac{1000\times6\times10^{-5}}{0.0015}=40\ \mathrm{S\,cm^2\,mol^{-1}}.

Degree of dissociation. [HX+]=cα=10−4[\ce{H+}]=c\alpha=10^{-4}, so α=10−41.5×10−3=115\alpha=\dfrac{10^{-4}}{1.5\times10^{-3}}=\dfrac1{15}.

Limiting value.
Λm∘=Λmα=40×15=600=6×102.\Lambda_m^\circ=\frac{\Lambda_m}\alpha=40\times15=600=6\times10^2.

Answer: 6

Topics:Conductance and Kohlrausch's lawpH and ionisation
← Previous questionNext question →