JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 18)
The resistance of a conductivity cell with cell constant 1.14 cm$$-$$1, containing 0.001 M KCl at 298 K is 1500 $$\Omega$$. The molar conductivity of 0.001 M KCl solution at 298 K in S cm2 mol$$-$$1 is ____________. (Integer answer)
Answer
760
Explanation
$$K = {1 \over R} \times {l \over A} = \left( {\left( {{1 \over {1500}}} \right) \times 1.14} \right)$$ S cm$$-$$1
$$ \Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$$ S cm2 mol$$-$$1
= 760 S cm2 mol$$-$$1
$$\Rightarrow$$ 760
$$ \Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$$ S cm2 mol$$-$$1
= 760 S cm2 mol$$-$$1
$$\Rightarrow$$ 760
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