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:

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A 5 m solution of H2SO4 is diluted from 1 litre to a volume of 100 litres, the normality of the solution will be

The correct option is C 0.1 NNormality = Molarity × n-factorn-factor of H2SO4=2 Molarity =MolesVolumes 5=Moles1 Moles = 5 Now. if volume will be =100 Then Molarity =5100 =0.05 M Now, normality =0.05×2 =0.1 N

A $5 m$ solution of $H_{2} S O_{4}$ is diluted from $1 l i t r e$ to a volume of $100 l i t r e s$ , the normality of the solution will be

The correct option is C $0.1 N$
Normality = Molarity $\times$ n-factor
n-factor of $H_{2} S O_{4} = 2$
Molarity $= \frac{Moles}{Volumes}$
$5 = \frac{Moles}{1}$
Moles = $5$
Now. if volume will be $= 100$
Then Molarity $= \frac{5}{100}$
$= 0.05 M$
Now, normality $= 0.05 \times 2$
$= 0.1 N$