What is the value of ${\mathrm{pK}}_{\mathrm{b}}<mspace\; width="1em"></mspace>({\mathrm{CH}}_3{\mathrm{COO}}^{-}<mspace\; width="1em"></mspace>)$ if ${\mathrm{\lambda }}_{\mathrm{m}}^0=<mspace\; width="1em"></mspace>390<mspace\; width="1em"></mspace>\&<mspace\; width="1em"></mspace>{\mathrm{\lambda }}_{\mathrm{m}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>7.8$ for 0.04 of a $<mspace\; width="1em"></mspace>{\mathrm{CH}}_3\mathrm{COOH}$  at $25°\mathrm{C}$ ?

(A) 9.3                                               

(B) 9.2

(C) 4.7                                               

(D) 4.8

At $25°C$, the standard emf of cell having reactions involving two electron change is found to be 0.295 V. The equilibrium constant of the reaction is-

$\left(A\right)<mspace\; width="1em"></mspace>29.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{-2}$

$\left(B\right)<mspace\; width="1em"></mspace>10<mspace\; width="1em"></mspace>$

$\left(C\right)<mspace\; width="1em"></mspace>{10}^{10}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

$\left(D\right)<mspace\; width="1em"></mspace>29.5<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^{10}$

How much current is necessary to produce $H_2$ gas at the rate of 1 $c{m}^3$ per second under STP ?

(A) 4.305 amp                                           

(B) 17.22 amp

(C) 8.61 amp                                            

(D) 2.1525 amp.

During discharge of a lead storage cell, the density of sulphuric acid in the cell-

1. Increases.                                     

2. Decreases.

3. Remains unchanged.                      

4. Initially increases but decreases subsequently.

$F_2$ gas can’t be obtained by the electrolysis of any $F^{-1}$salt because-

(1) Fluorine is the strongest reducing agent

(2) Fluorine is the strongest oxidising agent.

(3) Fluorine easily combine with atmospheric $O_2$

(4) All

The metal that cannot be produced on reduction of its oxide by aluminium is :

(1) K                                                        

(2) Mn

(3) Cr                                                      

(4) Fe

The calomel electrode is reversible with respect to-

1. Mercury                                       

2. $H^{+}$

3. $H{g}^{2+}$                                          

4. $C{l}^{-}$

The standard oxidation potentials of $Cu/C{u}^{2+}$and $C{u}^{+}<mspace\; width="1em"></mspace>/C{u}^{2+}<mspace\; width="1em"></mspace>are<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.34V<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.16<mspace\; width="1em"></mspace>V$respectively. The standard electrode potential of $C{u}^{+}<mspace\; width="1em"></mspace>/Cu$ would be :

1. 0.18V  

2. 0.52V

3. 0.82V    

4. 0.49V

 ${\mathrm{E}}^0<mspace\; width="1em"></mspace>\mathrm{for}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{reaction}<mspace\; width="1em"></mspace>$
$\mathrm{Fe}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Zn}}^{2+}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\mathrm{Zn}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Fe}}^{2+}<mspace\; width="1em"></mspace>\mathrm{is}<mspace\; width="1em"></mspace>–<mspace\; width="1em"></mspace>0.35<mspace\; width="1em"></mspace>\mathrm{V}.$
The   given   cell   reaction   is: 

1. Feasible  

2. Not feasible

3. In equilibrium 

4. None of the above

The  number of electrons involved in the deposition of 63.5 g of Cu from a solution of $CuS{O}_4$ is:

1 $6.022$ $\times$ ${10}^{23}$ $$

2 $3.011$ $\times$ ${10}^{23}$ $$

3 $12.044$ $\times$ ${10}^{23}$ $$

4 $$ $6.022$ $\times$ ${10}^{22}$