276072
The number of Faradays (F) required to produce $20 \mathrm{~g}$ of calcium from molten $\mathrm{CaCl}_{2}$ (Atomic mass of $\mathrm{Ca}^{2}=40 \mathrm{~mol}^{-1}$ ) is
1 1
2 2
3 3
4 4
Explanation:
Number of required moles - $\begin{aligned} & =\frac{\text { Mass }}{\text { Molar Mass }} \\ & =\frac{20}{40}=0.5 \mathrm{~mol} \end{aligned}$ Electricity required to produce 1 mole of calcium $=2 \mathrm{~F}$ The electricity required to produce $0.5 \mathrm{~mol}$ of $\mathrm{Ca}=0.5$ $\times 2 \mathrm{~F}$ $=1 \mathrm{~F}$
ELECTROCHEMISTRY
276073
During the electrolysis of molten sodium chloride, the time required to produce $0.10 \mathrm{~mol}$ of chlorine gas using a current of 3 amperes is
1 55 minutes
2 110 minutes
3 220 minutes
4 330 minutes
Explanation:
$\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $2 \mathrm{Cl}^{-} \underset{\text { electrolyse } 1 \text { mole }}{\longrightarrow} 1 \mathrm{Cl}_{2}+\underset{2 \mathrm{~F}}{2 \mathrm{e}^{-}}$ $\mathrm{Q}=\mathrm{It}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}$ For 1 mole of $\mathrm{Cl}_{2}$. produce $=2 \mathrm{~F}=2 \times 96500 \mathrm{C}$ Charge for 0.1 mole of $\mathrm{Cl}_{2}=2 \times 96500 \times 0.1$ Given $\mathrm{I}=3 \mathrm{~A}$ $\begin{aligned} & \therefore \mathrm{t}=\frac{2 \times 96500 \times 0.10}{3} \\ & \mathrm{t}=6433 \mathrm{sec} \text { or } \mathrm{t}=107.22 \mathrm{~min} \approx 110 \mathrm{~min} \end{aligned}$
NEET II-2016
ELECTROCHEMISTRY
276074
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = $1.60 \times 10^{-19} \mathrm{C}$ )
276075
When $0.1 \mathrm{~mol} \mathrm{MnO}_{4}^{2-}$ is oxidized, the quantity of electricity required to completely oxidise $\mathrm{MnO}_{4}^{2-}$ to $\mathrm{MnO}_{4}^{-}$is
1 $96500 \mathrm{C}$
2 $2 \times 96500 \mathrm{C}$
3 $9650 \mathrm{C}$
4 $96.50 \mathrm{C}$
Explanation:
The oxidation reaction $\begin{aligned} \mathrm{MnO}_{4}^{2-} & \longrightarrow \\ \mathrm{Q}=0.1 \times \mathrm{F} & =0.1 \times 96500 \mathrm{C} \\ & =9650 \mathrm{C} \end{aligned}$ The Quantity of electricity require to oxidizing $\mathrm{MnO}_{4}^{2-}$ completely will be 0.1 faraday i.e $9650 \mathrm{C}$.
276072
The number of Faradays (F) required to produce $20 \mathrm{~g}$ of calcium from molten $\mathrm{CaCl}_{2}$ (Atomic mass of $\mathrm{Ca}^{2}=40 \mathrm{~mol}^{-1}$ ) is
1 1
2 2
3 3
4 4
Explanation:
Number of required moles - $\begin{aligned} & =\frac{\text { Mass }}{\text { Molar Mass }} \\ & =\frac{20}{40}=0.5 \mathrm{~mol} \end{aligned}$ Electricity required to produce 1 mole of calcium $=2 \mathrm{~F}$ The electricity required to produce $0.5 \mathrm{~mol}$ of $\mathrm{Ca}=0.5$ $\times 2 \mathrm{~F}$ $=1 \mathrm{~F}$
ELECTROCHEMISTRY
276073
During the electrolysis of molten sodium chloride, the time required to produce $0.10 \mathrm{~mol}$ of chlorine gas using a current of 3 amperes is
1 55 minutes
2 110 minutes
3 220 minutes
4 330 minutes
Explanation:
$\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $2 \mathrm{Cl}^{-} \underset{\text { electrolyse } 1 \text { mole }}{\longrightarrow} 1 \mathrm{Cl}_{2}+\underset{2 \mathrm{~F}}{2 \mathrm{e}^{-}}$ $\mathrm{Q}=\mathrm{It}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}$ For 1 mole of $\mathrm{Cl}_{2}$. produce $=2 \mathrm{~F}=2 \times 96500 \mathrm{C}$ Charge for 0.1 mole of $\mathrm{Cl}_{2}=2 \times 96500 \times 0.1$ Given $\mathrm{I}=3 \mathrm{~A}$ $\begin{aligned} & \therefore \mathrm{t}=\frac{2 \times 96500 \times 0.10}{3} \\ & \mathrm{t}=6433 \mathrm{sec} \text { or } \mathrm{t}=107.22 \mathrm{~min} \approx 110 \mathrm{~min} \end{aligned}$
NEET II-2016
ELECTROCHEMISTRY
276074
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = $1.60 \times 10^{-19} \mathrm{C}$ )
276075
When $0.1 \mathrm{~mol} \mathrm{MnO}_{4}^{2-}$ is oxidized, the quantity of electricity required to completely oxidise $\mathrm{MnO}_{4}^{2-}$ to $\mathrm{MnO}_{4}^{-}$is
1 $96500 \mathrm{C}$
2 $2 \times 96500 \mathrm{C}$
3 $9650 \mathrm{C}$
4 $96.50 \mathrm{C}$
Explanation:
The oxidation reaction $\begin{aligned} \mathrm{MnO}_{4}^{2-} & \longrightarrow \\ \mathrm{Q}=0.1 \times \mathrm{F} & =0.1 \times 96500 \mathrm{C} \\ & =9650 \mathrm{C} \end{aligned}$ The Quantity of electricity require to oxidizing $\mathrm{MnO}_{4}^{2-}$ completely will be 0.1 faraday i.e $9650 \mathrm{C}$.
276072
The number of Faradays (F) required to produce $20 \mathrm{~g}$ of calcium from molten $\mathrm{CaCl}_{2}$ (Atomic mass of $\mathrm{Ca}^{2}=40 \mathrm{~mol}^{-1}$ ) is
1 1
2 2
3 3
4 4
Explanation:
Number of required moles - $\begin{aligned} & =\frac{\text { Mass }}{\text { Molar Mass }} \\ & =\frac{20}{40}=0.5 \mathrm{~mol} \end{aligned}$ Electricity required to produce 1 mole of calcium $=2 \mathrm{~F}$ The electricity required to produce $0.5 \mathrm{~mol}$ of $\mathrm{Ca}=0.5$ $\times 2 \mathrm{~F}$ $=1 \mathrm{~F}$
ELECTROCHEMISTRY
276073
During the electrolysis of molten sodium chloride, the time required to produce $0.10 \mathrm{~mol}$ of chlorine gas using a current of 3 amperes is
1 55 minutes
2 110 minutes
3 220 minutes
4 330 minutes
Explanation:
$\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $2 \mathrm{Cl}^{-} \underset{\text { electrolyse } 1 \text { mole }}{\longrightarrow} 1 \mathrm{Cl}_{2}+\underset{2 \mathrm{~F}}{2 \mathrm{e}^{-}}$ $\mathrm{Q}=\mathrm{It}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}$ For 1 mole of $\mathrm{Cl}_{2}$. produce $=2 \mathrm{~F}=2 \times 96500 \mathrm{C}$ Charge for 0.1 mole of $\mathrm{Cl}_{2}=2 \times 96500 \times 0.1$ Given $\mathrm{I}=3 \mathrm{~A}$ $\begin{aligned} & \therefore \mathrm{t}=\frac{2 \times 96500 \times 0.10}{3} \\ & \mathrm{t}=6433 \mathrm{sec} \text { or } \mathrm{t}=107.22 \mathrm{~min} \approx 110 \mathrm{~min} \end{aligned}$
NEET II-2016
ELECTROCHEMISTRY
276074
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = $1.60 \times 10^{-19} \mathrm{C}$ )
276075
When $0.1 \mathrm{~mol} \mathrm{MnO}_{4}^{2-}$ is oxidized, the quantity of electricity required to completely oxidise $\mathrm{MnO}_{4}^{2-}$ to $\mathrm{MnO}_{4}^{-}$is
1 $96500 \mathrm{C}$
2 $2 \times 96500 \mathrm{C}$
3 $9650 \mathrm{C}$
4 $96.50 \mathrm{C}$
Explanation:
The oxidation reaction $\begin{aligned} \mathrm{MnO}_{4}^{2-} & \longrightarrow \\ \mathrm{Q}=0.1 \times \mathrm{F} & =0.1 \times 96500 \mathrm{C} \\ & =9650 \mathrm{C} \end{aligned}$ The Quantity of electricity require to oxidizing $\mathrm{MnO}_{4}^{2-}$ completely will be 0.1 faraday i.e $9650 \mathrm{C}$.
276072
The number of Faradays (F) required to produce $20 \mathrm{~g}$ of calcium from molten $\mathrm{CaCl}_{2}$ (Atomic mass of $\mathrm{Ca}^{2}=40 \mathrm{~mol}^{-1}$ ) is
1 1
2 2
3 3
4 4
Explanation:
Number of required moles - $\begin{aligned} & =\frac{\text { Mass }}{\text { Molar Mass }} \\ & =\frac{20}{40}=0.5 \mathrm{~mol} \end{aligned}$ Electricity required to produce 1 mole of calcium $=2 \mathrm{~F}$ The electricity required to produce $0.5 \mathrm{~mol}$ of $\mathrm{Ca}=0.5$ $\times 2 \mathrm{~F}$ $=1 \mathrm{~F}$
ELECTROCHEMISTRY
276073
During the electrolysis of molten sodium chloride, the time required to produce $0.10 \mathrm{~mol}$ of chlorine gas using a current of 3 amperes is
1 55 minutes
2 110 minutes
3 220 minutes
4 330 minutes
Explanation:
$\mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$ $2 \mathrm{Cl}^{-} \underset{\text { electrolyse } 1 \text { mole }}{\longrightarrow} 1 \mathrm{Cl}_{2}+\underset{2 \mathrm{~F}}{2 \mathrm{e}^{-}}$ $\mathrm{Q}=\mathrm{It}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}$ For 1 mole of $\mathrm{Cl}_{2}$. produce $=2 \mathrm{~F}=2 \times 96500 \mathrm{C}$ Charge for 0.1 mole of $\mathrm{Cl}_{2}=2 \times 96500 \times 0.1$ Given $\mathrm{I}=3 \mathrm{~A}$ $\begin{aligned} & \therefore \mathrm{t}=\frac{2 \times 96500 \times 0.10}{3} \\ & \mathrm{t}=6433 \mathrm{sec} \text { or } \mathrm{t}=107.22 \mathrm{~min} \approx 110 \mathrm{~min} \end{aligned}$
NEET II-2016
ELECTROCHEMISTRY
276074
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = $1.60 \times 10^{-19} \mathrm{C}$ )
276075
When $0.1 \mathrm{~mol} \mathrm{MnO}_{4}^{2-}$ is oxidized, the quantity of electricity required to completely oxidise $\mathrm{MnO}_{4}^{2-}$ to $\mathrm{MnO}_{4}^{-}$is
1 $96500 \mathrm{C}$
2 $2 \times 96500 \mathrm{C}$
3 $9650 \mathrm{C}$
4 $96.50 \mathrm{C}$
Explanation:
The oxidation reaction $\begin{aligned} \mathrm{MnO}_{4}^{2-} & \longrightarrow \\ \mathrm{Q}=0.1 \times \mathrm{F} & =0.1 \times 96500 \mathrm{C} \\ & =9650 \mathrm{C} \end{aligned}$ The Quantity of electricity require to oxidizing $\mathrm{MnO}_{4}^{2-}$ completely will be 0.1 faraday i.e $9650 \mathrm{C}$.
