Question

The conductivity of $0.20M$ solution of $KCl$at $298K$ is $0.0248Sc{m^{ - 1}}$. Calculate its molar conductivity.
A.$112Sc{m^2}mo{l^{ - 1}}$
B.$124Sc{m^2}mo{l^{ - 1}}$
C.$136Sc{m^2}mo{l^{ - 1}}$
D.$148Sc{m^2}mo{l^{ - 1}}$

Answer 1

Hint:The formula of Molar conductivity is ${\Lambda _m} = \dfrac{{1000 \times k}}{M}$ OR conductance of one molar electrolyte. $k$ is conductivity of the electrolyte as given in the question.

Complete step by step answer:
The quantities given in question-
$k$(Conductivity) = $0.0248Sc{m^{ - 1}}$
$M$(molarity) = $0.20M$
$T$(Temperature) = $298K$
We have to find molar conductivity ${\Lambda _m}$
So the formula is ${\Lambda _m} = \dfrac{{1000 \times k}}{M}$
On putting values given in the problem,
${\Lambda _m} = \dfrac{{0.0248 \times 1000}}{{0.2}}$ $ \Rightarrow 124Sc{m^2}mo{l^{ - 1}}$
So the molar conductivity is $124Sc{m^2}mo{l^{ - 1}}$
-Conductance of metallic conductor decrease with temperature since resistance increase
Conductance of semiconductor and electrolyte increases with increasing temperature due to increased free number of electrons for semiconductor and increased mobility for electrolytes.
-Conductance of electrolyte is measured by passing alternative current since during passing direct current electrolysis occurs and denaturation takes place.
-In conductor-metric cells generally $KCl$ is used.
-Among all ions ${H^ + }$ have maximum conductance according to Grothus mechanism.
 -$k$ is specific conductance that is conductance of unit volume conductor and its formula is given as
$k = \dfrac{1}{\rho }$ , $\rho $ refers to resistivity
${10^{ - 3}}$On dilution conductance of the substance increases while its specific conductance decreases as the number of electrolytes per unit volume decreases but as a whole increases.
-Specific conductance can also be calculated by the below formula as it is the inverse of resistivity
$k = \dfrac{1}{R} \times \dfrac{l}{A}$ OR $k = G \times {G^*}$ , here $G$ represents conductance and ${G^*}$ represents cell
Cell constant ${G^*}$= $\dfrac{l}{A}$ , its unit is equal to ${m^{ - 1}}$ or $c{m^{ - 1}}$
Unit of $G = oh{m^{ - 1}}$ or mho or siemen(s)
Where $R$= resistance
$l$ = distance between electrodes
$A$= cross sectional area of electrode present in contact with electrolyte

Note:
Remember that the given formula should be used when specific conductance is given in $Sc{m^{ - 1}}$. If it is given in $S{m^{ - 1}}$ then at the place of $1000$ there will be ${10^{ - 3}}$ as shown below,
${\Lambda _m} = \dfrac{{k \times {{10}^{ - 3}}}}{M}$