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NCERT Class XII Chemistry
Chapter - Electrochemistry
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Question : 8 of 32
Marks: +1, -0
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1^{-1}. Calculate its molar conductivity.
Solution:  
Given κ=2.48×10−2 W−1 cm−1,\kappa=2.48 \times 10^{-2} \ \mathrm{W}^{-1} \ \mathrm{cm}^{-1},
M=0.20 mol L−1M=0.20 \ \mathrm{mol} \ \mathrm{L}^{-1}
λm=κ×1000M\lambda_{m}=\frac{\kappa \times 1000}{M}
=1000×2.48×10−20.20=\frac{1000 \times 2.48 \times 10^{-2}}{0.20}
=124 Ω−1 cm2 mol−1=124 \ \Omega^{-1} \ \mathrm{cm}^{2} \ \mathrm{mol}^{-1}
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