The specific conductance of the saturated solution of AgCl is $1.2<mspace\; width="1em"></mspace>\times {10}^{-6}<mspace\; width="1em"></mspace>{\Omega }^{-1}<mspace\; width="1em"></mspace>c{m}^{-1}$. Molar conductance at infinite dilution of AgCl is $119<mspace\; width="1em"></mspace>{\Omega }^{-1}<mspace\; width="1em"></mspace>c{m}^2<mspace\; width="1em"></mspace>mo{l}^{-1}$. The solubility of AgCl in mole/L is

1.  $0.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>M$

2.  $1.04<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>M$

3.  $1.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>M$

4.  $2<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>M$

How many coulombs of electric charge must pass through acidulated water in order to release 22.4 litre of H₂ at N.T.P?

1  65900 coulomb

2  193000 coulomb

3  96500 coulomb

4  96500 faraday

For HCl solution at $25°\mathrm{C}$, equivalent conductance at infinite dilution, is $425<mspace\; width="1em"></mspace>{\mathrm{ohm}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{cm}}^2<mspace\; width="1em"></mspace>{\mathrm{equiv}}^{-1}$. The specific conductance of a solution of HCl is $0.3825<mspace\; width="1em"></mspace>{\mathrm{ohm}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{cm}}^{-1}$. If the apparent degree of dissociation is 90% the normality of the solution is:-

1. 0.90 N

2. 1.0 N

3. 10 N

4. 1.2 N

The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca = 40 g mol^–1) is:

1. 2

2. 3

3. 4

4. 1

One liter of 0.5 M KCl solution is electrolyzed for one minute in a current of 1.608 mA. Considering 100 % efficiency, the pH of the resulting solution will be : 

Equivalent conductance of saturated ${\mathrm{BaSO}}_4$ solution is 400 ${\mathrm{ohm}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{cm}}^{2<mspace\; width="1em"></mspace>}{\mathrm{equivalent}}^{-1}$ and it's specific conductance is $8<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-5}<mspace\; width="1em"></mspace>{\mathrm{ohm}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{cm}}^{-1}$; hence solubility product ${\mathrm{K}}_{\mathrm{sp}}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>{\mathrm{BaSO}}_4$ is : 

1.  $4<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-8}<mspace\; width="1em"></mspace>{\mathrm{M}}^2$

2.  $1<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-8}<mspace\; width="1em"></mspace>{\mathrm{M}}^2$

3.  $2<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-4}<mspace\; width="1em"></mspace>{\mathrm{M}}^2$

4.  $1<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-4}<mspace\; width="1em"></mspace>{\mathrm{M}}^2$

The specific conductance of 0.01 M solution of a weak monobasic acid is 0.20 x 10^-3 S cm^-1. The dissociation constant of the acid is-

[Given  ${\Lambda }_{\mathrm{HA}}^{\infty }$ = 400 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is $1.65<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-4}$ S ${\mathrm{cm}}^{-1}$. The degree of dissociation of acetic acid is [Given: Equivalent conductance at infinite dilution of ${\mathrm{H}}^{+}$ = 349.1 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$ and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ = 40.9 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$)

1.  0.021

2.  0.21

3.  0.012

4.  0.12

$$
For the Cell \(Pt(s)|| Br^-(aq)(0.010M)| Br_2(l)|| H^+(aq) (0.0030M) |H_2(g) (1 bar) | Pt(s)\)

If the concentration of ${\mathrm{Br}}^{-}$ becomes 2 times and the concentration of ${\mathrm{H}}^{+}$ becomes half of the initial value, then emf of the cell

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same

Adding powdered lead and iron to a solution that is 1.0 molar each in ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ ions, would result to a reaction in which

     $\begin{array}{rl}& {\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\circ }=-0.44\mathrm{V}\\ & {\mathrm{E}}_{{\mathrm{Pb}}^{2+}/\mathrm{Pb}}^0=-0.13\mathrm{V}\end{array}$

1.  Conc. of both ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Fe}}^{2+}$ are increased

2.  More lead and iron are formed

3.  More of iron and ${\mathrm{Pb}}^{2+}$ ions are formed

4.  More of lead and ${\mathrm{Fe}}^{2+}$ ions are formed