Verified Solution • 2025–26 Session moderate

QJEEM14Ch303f: JEE Main 2014 Chemistry - Class 12 JEEM14Ch303f Solution: Resistance of 0.2 M solution of an electrolyte is 50 . The…

Resistance of 0.2 M solution of an electrolyte is 50 Ω\Omega . The specific conductance of the solution is 1.4 S m-1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω\Omega . The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is :

Multiple Choice Options

  • A 5 × 103
  • B 5 × 102
  • C 5 × 10-4
  • D 5 × 10-3

Step-by-Step Verified Solution

Given for 0.20.2 MM solution

R=50ΩR = 50\Omega

κ=1.45 S m−1=1.4×10−2 S cm−1\kappa = 1.45\,S\,{m^{ - 1}} = 1.4 \times {10^{ - 2}}\,S\,c{m^{ - 1}}

Now, R=ρℓa=1κ×ℓaR = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}

⇒ℓa=R×κ=50×1.4×10−2\Rightarrow {\ell \over a} = R \times \kappa = 50 \times 1.4 \times {10^{ - 2}}

For 0.50.5 MM solution

R=280 Ω; κ=?R = 280\,\Omega ;\,\kappa = ?

ℓa=50×1.4×10−2{\ell \over a} = 50 \times 1.4 \times {10^{ - 2}}

⇒R=ρℓa=1κ×ℓa\Rightarrow R = \rho {\ell \over a} = {1 \over \kappa } \times {\ell \over a}

⇒    κ=1280×50×1.4×10−2\Rightarrow \,\,\,\,\kappa = {1 \over {280}} \times 50 \times 1.4 \times {10^{ - 2}}

=1280×70×10−2=2.5×10−3 S cm−1= {1 \over {280}} \times 70 \times {10^{ - 2}} = 2.5 \times {10^{ - 3}}\,S\,c{m^{ - 1}}

Now, Λm=κ×1000M=2.5×10−3×10000.5{\Lambda _m} = {{\kappa \times 1000} \over M} = {{2.5 \times {{10}^{ - 3}} \times 1000} \over {0.5}}

=5 S cm2mol−1=5×10−4 S m2 mol−1= 5\,S\,c{m^2}mo{l^{ - 1}} = 5 \times {10^{ - 4}}\,S\,{m^2}\,mo{l^{ - 1}}

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