Please be patient as the PDF generation may take upto a minute.

One gm metal $M^{3+}$ was discharged by the passage of 1.81 x ${10}^{23}$ electrons.
What is the atomic weight of metal?
1. 33.35
2. 133.4
3. 66.7
4. None of these

How much time is required for complete decomposition of 4 moles of water using 4 ampere?
1. 3.86 X ${10}^5$ sec
2. 1.93 X ${10}^5$ sec
3. 96500 sec
4. 48250 sec

If 0.50 L of a 0.60 M SnS$O^4$ solution is electrolyzed for a period of 30.0 min
using a current of 4.60 A. If inert electrodes are used, what is the final
concentration of $S{n}^{2+}$ remaining in the solution ? [at. Wt. of Sn = 119]
1. 0.342 M
2. 0.544 M
3. 0.389 M
4. 0.514 M

An electrolysis of a oxytungsten complex ion using 1.10 A for 40 min produces 0.838 g of tungsten. What is the charge on tungsten in the material ? (Atomic weight: W = 184)
1. 6
2. 2
3. 4
4. 1

Cadmium amalgam is prepared by electrolysis of a solution of Cd$C{l}_2$ using a
mercury cathode. How long should a current 4 A be passed in order to prepare 10% by wt. Cd in Cd-Hg amalgam on cathode of 4.5 g Hg? (atomic weight of Cd = 112)
1. 400 sec
2. 215.40 sec
3. 861.6 sec
4. 430.8 sec

$l_2$ (s)| l $l^{-}$  (0.1 M) half cell ¡s connected to a $H^{+}$ (aq) | $H_2$ (1 bar) | Pt half cell and emf is found to be 0.7714 V. If $E_{I_2/I^{-}}^0$ =  0.535 V, find the pH of $H^{+}| H_2$ half cell.
1. 1
2. 3
3. 5
4. 7

 If ${E^0}_{A{u}^{+}/Au}$ is 1.69 V and ${E^0}_{A{u}^{3+}/Au}$ is 1.40 V, then ${E^0}_{A{u}^{+3}/A{u}^{1+}}$ will be :
1. 0.19 V
2. 2.945 V
3. 1.255 V
4. None of these

Consider the following standard electrode potentials and calculate the
equilibrium constant at 25°C for the indicated disproportionation reaction :
$3M{n}^{2+}\left(aq\right)\to Mn\left(s\right)+2M{n}^{3+}aq$
$M{n}^{3+}\left(aq\right)+e^{-}\to M{n}^{2+}\left(eq\right)$; $E^0 = 1.51 V$
$M{n}^{2+}\left(aq\right)+2e^{-}\to Mn\left(s\right)\left(eq\right)$; $E^0 = -1.185 V$
1. 1.2 X ${10}^{-43}$
2. 2.4 X ${10}^{-73}$
3. 6.3 X ${10}^{-92}$
4. 1.5 X ${10}^{-62}$

Ionisation constant of a weak acid (HA) in terms of ${\wedge }_m^{\infty }$ and ${\vee }_m$ is
1. $K_{\alpha }$ =  $\frac{C{\wedge }_m^{\infty }}{{\wedge }_m-{\wedge }_m^{\infty }}$
 
2. $K_{\alpha }$ = $\frac{C{\wedge }_m^{\infty }}{{\wedge }_m^{\infty }\left({\wedge }_m^{\infty }-{\wedge }_m^{}\right)}$
 
3.$K_{\alpha }$= $\frac{C{\left({\wedge }_m^{\infty }\right)}^2}{{\wedge }_m^{\infty }-{\wedge }_m^{}}$
 
4. None of these

Equivalent conductivity of ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$is related to molar conductivity by the expression:
1. ${\wedge }_{eq}$ = ${\wedge }_m$
2. ${\wedge }_{eq}$ = ${\wedge }_m/3$
3. ${\wedge }_{eq}$ = $3{\wedge }_m$
4. ${\wedge }_{eq}$ = ${\wedge }_m/6$

The limiting equivalent conductivity of NaCl, KCl and KBr are 126.5, 150.0 and
151.55 $c{m}^2$ $e{q}^{-1}$, respectively. The limiting equivalent ionic conductance for $B{r}^{-}$ is 78 S $c{m}^2$ $e{q}^{-1}$. The limiting equivalent ionic conductance for $N{a}^{+}$ ions would be :
1. 128
2. 125
3. 49
4. 50

The increase in equivalent conductance of a weak electrolyte with dilution is
due to 
1. Increase in degree of dissociation and decrease ¡n ionic mobility
2. Decrease in degree of dissociation and decrease in ionic mobility
3. Increase in degree of dissociation and increase in ionic mobility
4. Decrease in degree of dissociation and increase in ionic mobility

The electric conduction of a salt solution in water depends on the:
1. Size of its molecules
2. Shape of its molecules
3. Size of solvent molecules
4. Extent of its ionization

A graph was plotted molar conductivity of various electrolytes (NaCl, HCl and
$N{H}_4$OH) and $\sqrt{C}$(in $mo{l}^{-1}$). Correct setting :

1. l(NaCl), ll(HCl), lll($N{H}_4$OH)
2. l(HCl), ll(NaCl), lll($N{H}_4$OH)
3. l(N$N{H}_4$OH), ll(NaCl), lll(HCl)
4. l($N{H}_4$OH), ll(HCl), lll(NaCl)