Fully charged lead storage battery contains $1.5 L$ of $5 M H_{2} S O_{4}$ . If $2.5 a m p$ of current is taken from the cell for $965$ minutes, then what will be the molarity of remaining $H_{2} S O_{4}$ ? Assume that volume of battery fluid to be constant.

4 M

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3.5 M

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2 M

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4.25 M

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Solution

The correct option is A $4 M$
$i n l e a d s t o r a g e b a t t r y$
$m a s s o f P b d e p o s i t e d$
$w = Z i t = \frac{m o l . w t .}{n F} \times i t$
$w = \frac{207.2 \times 2.5 \times 965 \times 60}{1 \times 96500}$
$w = 310.8 g$
$m o l e o f P b d e p o s i t e d = \frac{310.8}{207.2} = 1.5 m o l e$
$t o t a l m o l e o f s u l p h u r i c a c i d = 5 \times 1.5 = 7.5 m o l e$
$m o l e o f s u l p h u r i c a c i d i n s o l u t i o n = 7.5 - 1.5 = 6 m o l e$
$m o l a r i t y o f s u l p h u r i c a c i d = \frac{6}{1.5 L} = 4 M$

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A lead storage battery containing 5.0L of $H_{2} S O_{4}$ solution is operated for $9.65 \times 10^{5} s$ with a steady current of 100 mA. Assuming volume of the solution remains constant, normality of $H_{2} S O_{4}$ will :

A lead storage battery containing 5.0 L of (1N) $H_{2} S O_{4}$ solution is operated for $9.65 \times 10^{5}$ s with a steady current of 100 mA. Assuming volume of the solution remaining constant, normality of $H_{2} S O_{4}$ will:

During discharging of lead-storage acid battery following reaction takes place:

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If 2.5 amp of current is drawn for 965 minutes, $H_{2} S O_{4}$ consumed is:

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Fully charged lead storage battery contains 1.5 L of 5 M H2SO4. If 2.5 amp of current is taken from the cell for 965 minutes, then what will be the molarity of remaining H2SO4? Assume that volume of battery fluid to be constant.

The correct option is A 4 MinleadstoragebattrymassofPbdepositedw=Zit=mol.wt.nF×itw=207.2×2.5×965×601×96500w=310.8gmoleofPbdeposited=310.8207.2=1.5moletotalmoleofsulphuricacid=5×1.5=7.5molemoleofsulphuricacidinsolution=7.5−1.5=6molemolarityofsulphuricacid=61.5L=4M

Fully charged lead storage battery contains $1.5 L$ of $5 M H_{2} S O_{4}$ . If $2.5 a m p$ of current is taken from the cell for $965$ minutes, then what will be the molarity of remaining $H_{2} S O_{4}$ ? Assume that volume of battery fluid to be constant.