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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
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)

## 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
2

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
For the galvanic cell,

Zn(s) + Cu2+ (0.02 M) $$\to$$ Zn2+ (0.04 M) + Cu(s),

Ecell = ______________ $$\times$$ 10$$-$$2 V. (Nearest integer)

[Use : $$E_{Cu/C{u^{2 + }}}^0$$ = $$-$$ 0.34 V, $$E_{Zn/Z{n^{2 + }}}^0$$ = + 0.76 V, $${{2.303RT} \over F} = 0.059\,V$$]

## Explanation

Galvanic cell :

$$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$$

Nernst equation = $${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$$

$$\Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$$

$$\Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$$

$$\Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$$

= 1.0911 = 109.11 $$\times$$ 10$$-$$2

= 109
3

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
For the cell

Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)

the cell potential E1 = 0.3095 V

For the cell

Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)

the cell potential = ____________ $$\times$$ 10$$-$$2 V. (Round off the nearest integer).

[Use : $${{2.303RT} \over F}$$ = 0.059]

## Explanation

Cell reaction is :

$$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$$

Now, $${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$ .... (1)

$$\therefore$$ $${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$$ ....(2)

From (1) and (2), E2 = 0.28 V = 28 $$\times$$ 10$$-$$2 V
4

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
The conductivity of a weak acid HA of concentration 0.001 mol L$$-$$1 is 2.0 $$\times$$ 10$$-$$5 S cm$$-$$1. If $$\Lambda _m^o$$(HA) = 190 S cm2 mol$$-$$1, the ionization constant (Ka) of HA is equal to ______________ $$\times$$ 10$$-$$6. (Round off to the Nearest Integer)

## Explanation

$$\Lambda _m^{} = 1000 \times {\kappa \over M}$$

$$= 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$$ S cm2 mol$$-$$1

$$\Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$$

HA $$\rightleftharpoons$$ H+ + A$$-$$

$$0.001(1 - \alpha )0.001\alpha 0.001\alpha$$

$$\Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$$

$$= 12.3 \times {10^{ - 6}}$$

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