JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 16)
A 5.0 m mol dm$$-$$3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm$$-$$1. The molar conductivity of this solution is ___________ mSm2 mol$$-$$1. (Round off to the Nearest Integer).
Answer
14
Explanation
Conductance = $${{Conductivity} \over {Cell\,cons\tan t}}$$
$$ \therefore $$ Conductivity = 0.55 $$\times$$ 10$$-$$3 $$\times$$ 1.3 S cm$$-$$1
Molar conductivity = $${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$$
$$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$$
= 143 S cm2 mol$$-$$1
= 14.3 mS m2 mol$$-$$1
$$ \approx $$ 14 mS m2 mol$$-$$1
$$ \therefore $$ Conductivity = 0.55 $$\times$$ 10$$-$$3 $$\times$$ 1.3 S cm$$-$$1
Molar conductivity = $${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$$
$$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$$
= 143 S cm2 mol$$-$$1
= 14.3 mS m2 mol$$-$$1
$$ \approx $$ 14 mS m2 mol$$-$$1
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