How much volume of 1 ,N aqueous solution of H | vedclass

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(B) The normality $(N)$ of a solution is related to its molarity $(M)$ by the formula: $N = M 	imes 	ext{n-factor}$.For $H_2SO_4$, the n-factor (bas

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$400$

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(B) The normality $(N)$ of a solution is related to its molarity $(M)$ by the formula: $N = M 	imes 	ext{n-factor}$.For $H_2SO_4$, the n-factor (basicity) is $2$.Given $N = 1 \,N$, we have $1 = M 	imes 2$, which implies $M = 0.5 \,M$.We know that Molarity $(M)$ is defined as the number of moles $(n)$ per volume $(V)$ in liters: $M = \frac{n}{V(L)}$.Substituting the given values: $0.5 = \frac{0.2}{V(L)}$.Solving for $V(L)$: $V(L) = \frac{0.2}{0.5} = 0.4 \,L$.Converting to milliliters: $0.4 \,L 	imes 1000 \,mL/L = 400 \,mL$.

How much volume of $1 \,N$ aqueous solution of $H_2SO_4$ should be taken, which will contain $0.2$ moles of $H_2SO_4$ (in $\,mL$)?

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