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Standard XII

Chemistry

Charge and Current

The amount of...

Question

The amount of charge which the battery must have been used is:

9.88 F

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8.98 F

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8.89 F

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7.88 F

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Solution

The correct option is A $9.88 F$
wt. of $H_{2} S O_{4}$ at initial = 3.51000.391.294 = 1766.31 gm
wt. of $H_{2} S O_{4}$ after decay =3.510001.30.20 = 797.3 gm
amount deposite = 969 gm
n-fector of $H_{2} S O_{4}$ = 1
w = zQ
Q = 969.0F $\frac{1}{98}$
Q = 9.88F

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Similar questions

Q. The number of amphere-hours for which the battery must have been used is:

Q. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139

g m L^{- 1}

H_{2} S O_{4}

of density

1.294 g m L^{- 1}

is 39% and that of density

1.139 g m L^{- 1}

is 20% by weight.The battery holds 3.5 L of acid and the volume practically remains constant during discharge. The number of ampere hours for which the battery must have been used are........................
The discharging and charging reactions are:

P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-}

(charging)

P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O

(discharging)

Q. The two parallel plates of a condenser have been connected to a battery of

300 V

and the charge collected at each plates is

1 μ C

. The energy supplied by the battery is :

During the discharge of a lead storage battery, the density of sulphuric acid fell from $1.294$ to $1.139$ gm $L^{- 1}$ . Sulphuric acid of density $1.294$ gm $L^{- 1}$ is $39$ % by weight and that of density 1.139 gm $L^{- 1}$ is $20$ % by weight. The battery holds $3.5$ litres of acid and the column practically remained constant during the discharge.

Calculate the number of ampere-hours for which the battery must have been used. The charging and discharging reactions are :

$P b + S O {_{4}}^{2 -} \to P b S O_{4} + 2 e^{-}$ (charging)

$P b O_{2} + 4 H^{+} + S O {_{4}}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O$ (discharging)

Q. A parallel-plate capacitor having plate area

25 c m^{2}

and separation

1.00 m m

is connected to a battery of

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Calculate the charge flown through the battery. How much work has been done by the battery during the process?

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The amount of charge which the battery must have been used is:

The correct option is A 9.88 Fwt. of H2SO4 at initial = 3.5*1000*.39*1.294 = 1766.31 gmwt. of H2SO4 after decay =3.5*1000*1.3*0.20 = 797.3 gmamount deposite = 969 gmn-fector of H2SO4 = 1w = zQ Q = 969.0F198 Q = 9.88F

The correct option is A $9.88 F$
wt. of $H_{2} S O_{4}$ at initial = 3.51000.391.294 = 1766.31 gm
wt. of $H_{2} S O_{4}$ after decay =3.510001.30.20 = 797.3 gm
amount deposite = 969 gm
n-fector of $H_{2} S O_{4}$ = 1
w = zQ
Q = 969.0F $\frac{1}{98}$
Q = 9.88F