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How many second would it take to produce enough aluminum by the Hall process to make a case of 24 cans of aluminum soft-drink, if each can uses 5.0 g of Al, a current of 9650 amp is employed, and the current efficiency of the cell is 90.0%:
(1) 203.2
(2) 148.14
(3) 333
(4) 6.17

108 g fairly concentrate solution of $AgN{O}_3$ is electrolyzed using 0.1 F of electricity. The weight of resulting solution is:
(1) 94 g
(2) 11.6 g
(3) 96.4 g
(4) None of these

The electrolysis of acetate solution produces ethane according to reaction:
$2C{H}_{3}CO{O}^{-}\to C_2{H}_6\left(g\right)+2C{O}_2\left(g\right)+2e^{-}$
The current efficiency of the process is 80%.  What volume of gases would be produced at 27$°$C and 740 torr, if the current of 0.5 amp is passed through the solution for 96.45 min?
(1) 6.0 L                   (2) 0.60 L                   (3) 1.365 L                       (4) 0.91 L

100 mL of 0.05 M $CuS{O}_4\left(aq\right)$ solution was electrolyzed using inert electrodes by passing current till the pH of the resulting solution was 2.  The solution after electrolysis was neutralized and then treated with excess KI and formed $I_2$ titrated with 0.04 M $N{a}_2{S}_2{O}_3$. Calculate the required volume (in mL) of $N{a}_2{S}_2{O}_3:$
(1) 112.5 mL
(2) 100 mL
(3) 125 mL
(4) None of these

If the equilibrium constant for the reaction $H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\rightleftharpoons H_{2}O\left(l\right)$ is ${10}^{13}$ at certain temperature then what is the $E^{°}$ for the reaction, $2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$ $Given: \frac{2.303RT}{F}=0.066$
(1) 1.230 V
(2) -0.858 V
(3) -0.80 V
(4) -0.8274 V

A fuel cell develops an electrical potential from the coombustion of butane at 1 bar and 298 K
$C_4{H}_{10}\left(g\right)+6.5O_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(l\right);$
${∆}_r{G}^{°}=-2746 kJ/mol$
What is $E^{°}$ of a cell?
(1) 4.74 V
(2) 0.547 V
(3) 4.37 V
(4) 1.09 V

Given the cell: \(\mathrm{Cd}(s)\left|\mathrm{Cd}(\mathrm{OH})_2(s)\right| \mathrm{NaOH}(a q, 0.01 M) \mid H_2(g), 1 \text { bar } \mid P t(s)\)
with E_cell = 0.0 V.  If E°_Cd²⁺lCd = -0.39 V, then Ksp  of  Cd(OH)₂ is:
1. 0.1
2. ${10}^{-13}$
3. ${10}^{-15}$
4. None of these

Calculate the potential of a half cell having reaction :
$A{g}_{2}S\left(s\right)+2e^{-}\rightleftharpoons 2Ag\left(s\right)+S^{2-}\left(aq\right)$
in a solution buffered at pH = 3 and which is also saturated with 0.1 $M{H}_{2}S$(aq):
$[Given: K_{sp}(A{g}_{2}S)={10}^{-49}, K_{a_1}·K_{a_1}={10}^{-21}]$
(1) 1.18
(2) 0.19
(3) -0.19 V
(4) none of these

The conductivity of 0.1 N NaOH solution is 0.022 S $c{m}^{-1}$.  When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreases to 0.0055 S $c{m}^{-1}$.  The equivalent conductivity in S $C{m}^2$ $equivalen{t}^{-1}$ of NaCl solution is :
(1) 0.0055
(2) 0.11
(3) 110
(4)None of these
 

In above question after formation of NaCl, further 0.1 N HCl is added, the volume of which is double to that of the first portion added, the conductivity increase to 0.018 S $c{m}^{-1}$.  The value of ${\wedge }_{eq}$(HCl) is [assume no change in conductivity of NaCl (aq)]:
1. 330 S $c{m}^2 e{q}^{-1}$
2. 305 S $c{m}^2 e{q}^{-1}$
3. 415 S $c{m}^2 e{q}^{-1}$
4. 360 S $c{m}^2 e{q}^{-1}$

Given the following molar conductivities at ${25}^{°}C:, HCl 426 {\Omega }^{-1} c{m}^2 mo{l}^{-1};$
$$$NaCl 126 {\Omega }^{-1} c{m}^2 mo{l}^{-1}; NaC (sodium coronate)$$83 {\Omega }^{-1} c{m}^2 mo{l}^{-1}$.  what is the ionization constant of crotonic acid? If the conductivity of a 0.001 M crotonic acid solution is 3.83$\times$${10}^{-5} {\Omega }^{-1}c{m}^{-1}$?
(1) ${10}^{-5}$                 (2) $1.11\times {10}^{-5}$
(3) $1.11\times {10}^{-4}$      (4)0.01

Equivalent conductivity of $BaC{l}_2, H_{2}S{O}_4 and HCl,$ are $x_1, x_2 and x_3 S c{m}^{-1} e{q}^{-1}$ at infinite dilution.  If conductivity of saturated $BaS{O}_4$ solution is x S $c{m}^{-1}$, then $K_{sp} of BaS{O}_4$ is:
(1) $\frac{500x}{(x_1+x_2-2x_3)}$                         (2) $\frac{{10}^6{x}^2}{{(x}^{}}$₁+x₂-2x₃)3
(3) $\frac{2.5\times {10}^5{x}^2}{{(x}^{}}$₁+x₂-x₃)2                          (4) $\frac{0.25 x^2}{{(x}^{}}$₁+x₂-2x₃)2

The conductivity of 0.001 M $N{a}_{2}S{O}_4$ solution is $2.6\times {10}^{-4} Sc{m}^{-1}$ and increase to $7.0\times {10}^{-4} Sc{m}^{-1}$, when the solution is saturated with $CaS{O}_4$. The molar conductivities of $N{a}^{+}$ and $C{a}^{2+}$ are 50 and 120 $Sc{m}^{2}mo{l}^{-1}$, respectively.  Neglect conductivity of used water. What is the solubility product of $CaS{O}_4$?
(1) $4\times {10}^{-6}$                     (2) $1.57\times {10}^{-3}$
(3) $4\times {10}^{-4}$                     (4) $2.46\times {10}^{-6}$
     

The ionization constant of a weak acid is $1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $380\times {10}^{-4}$ S$m^{2}mo{l}^{-1}$.  If the cell constant is 0.01 $m^{-1}$ then conductance of 0.01 M acid solution is:
(1) $1.52\times {10}^{-5} S$
(2) 1.52 S
(3) 1.52 $\times {10}^{-3}$ S
(4) $1.52\times {10}^{-4} S$