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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 10 of 30
Marks: +1, -0
The conductivity of 0.20M0.20 \mathrm{M} solution of KCl\mathrm{KCl} at 298K298 \mathrm{K} is 0.025Scm−10.025 \mathrm{S} \mathrm{cm}^{-1}, Calculate its molar conductivity.
Solution:  
  Conductivity    =0.025Scm−1\; \text{Conductivity} \;\;=0.025 \mathrm{S} \mathrm{cm}^{-1}
  molarity,  M  =0.20M\; \text{molarity,}\; \mathrm{M}\;=0.20 \mathrm{M}
T=298KT =298 \mathrm{K}
Tm=?T_m =?
molar conductivity =?=?
Tm=κ×1000CT_m = \frac{\kappa \times 1000}{C}
=0.025×1000Scm−10.20molcm−3= \frac{0.025 \times 1000 \mathrm{S} \mathrm{cm}^{-1}}{0.20 \mathrm{mol} \mathrm{cm}^{-3}}
=250.20= \frac{25}{0.20}
=125Scm2mol−1=125 \mathrm{S} \mathrm{cm}^2 \mathrm{mol}^{-1}
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