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Electrochemistry question

2022 · 25 Jul · Shift 2 · Q4

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Electrochemistry
2022 · 25 Jul · Shift 2 · Q4

Electrochemistry question

2022 · 25 Jul · Shift 2 · Q4

JEE MainChemistryElectrochemistryMCQ+4 / −1

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is

{\Lambda _{m1}}

and that of 20 moles another identical cell heaving 80 mL NaCl solution is

{\Lambda _{m2}}

. The conductivities exhibited by these two cells are same. The relationship between

{\Lambda _{m2}}

and

{\Lambda _{m1}}

is

A
${\Lambda _{m2}}$ = 2 ${\Lambda _{m1}}$
B
${\Lambda _{m2}}$ = ${\Lambda _{m1}}$ / 2
C
${\Lambda _{m2}}$ = ${\Lambda _{m1}}$
D
${\Lambda _{m2}}$ = 4 ${\Lambda _{m1}}$

View written solutionFree

Correct answer: A

Use the relation between conductivity and molar conductivity

For an electrolyte solution,

\Lambda_m = \kappa \frac{1000}{C}

where:

$\Lambda_m$ = molar conductivity
$\kappa$ = conductivity
$C$ = concentration in mol L $^{-1}$

So,

\Lambda_m \propto \frac{\kappa}{C}

Given that the conductivities of the two cells are same, we have

\kappa_1 = \kappa_2

Hence,

\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{C_1}{C_2}

Find concentration of first solution

First cell contains $10$ moles in $20$ mL.

Convert volume to litre:

20\text{ mL} = 0.020\text{ L}

Thus,

C_1 = \frac{10}{0.020} = 500\,\text{mol L}^{-1}

Find concentration of second solution

Second cell contains $20$ moles in $80$ mL.

Convert volume to litre:

80\text{ mL} = 0.080\text{ L}

Thus,

C_2 = \frac{20}{0.080} = 250\,\text{mol L}^{-1}

Compare molar conductivities

Since $\kappa_1 = \kappa_2$ ,

\frac{\Lambda_{m2}}{\Lambda_{m1}} = \frac{C_1}{C_2} = \frac{500}{250} = 2

Therefore,

\Lambda_{m2} = 2\Lambda_{m1}

Option check

A: $\Lambda_{m2} = 2\Lambda_{m1}$ ✅
B: $\Lambda_{m2} = \Lambda_{m1}/2$ ❌
C: $\Lambda_{m2} = \Lambda_{m1}$ ❌
D: $\Lambda_{m2} = 4\Lambda_{m1}$ ❌

Therefore, the correct option is A.

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