(B) The molarity $(M)$ of $H_2SO_4$ is $5 \ M$. The n-factor for $H_2SO_4$ is $2$. Normality $(N)$ = Molarity $(M)$ $\times$ n-factor = $5 \times 2 =

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(B) The molarity $(M)$ of $H_2SO_4$ is $5 \ M$. The n-factor for $H_2SO_4$ is $2$. Normality $(N)$ = Molarity $(M)$ $\times$ n-factor = $5 \times 2 = 10 \ N$. Using the dilution formula $N_1V_1 = N_2V_2$: $10 \ N \times 1 \ L = N_2 \times 10 \ L$. $N_2 = \frac{10 \times 1}{10} = 1 \ N$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

Easy

What will be the normality of a $1 \ L$ solution of $5 \ M$ $H_2SO_4$ after diluting it to $10 \ L$ by adding water (in $N$)?

A
$0.5$
B
$1.0$
C
$5.0$
D
$10.0$

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Method of expressing concentration of solution

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What will be the normality of a $1 \ L$ solution of $5 \ M$ $H_2SO_4$ after diluting it to $10 \ L$ by adding water (in $N$)?

Similar Questions

If $22 \ g$ of benzene $(C_6H_6)$ is dissolved in $122 \ g$ of carbon tetrachloride $(CCl_4)$,calculate the mass percentage of benzene. (in $\%$)

The strength of a solution $(S)$ in $g/L$ is related to its normality $(N)$ and equivalent weight $(E)$ of the solute by the formula:

$A$ $1$ molal solution is defined as $1$ mole of solute dissolved in:

The density of a $10.0 \% \, w/w$ aqueous solution of $KCl$ is $1.06 \, g \, cm^{-3}$. The molarity of this solution is .........

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