2.5 cm^{3} of 0.2 M H_{2}SO_{4} solution i | vedclass

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(C) Given: Initial volume $V_{1} = 2.5 \ cm^{3} = 2.5 	imes 10^{-3} \ dm^{3}$.Initial molarity $M_{1} = 0.2 \ M$.Since $H_{2}SO_{4}$ is a dibasic aci

Vedclass · Accepted Answer

$0.002$

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(C) Given: Initial volume $V_{1} = 2.5 \ cm^{3} = 2.5 	imes 10^{-3} \ dm^{3}$.Initial molarity $M_{1} = 0.2 \ M$.Since $H_{2}SO_{4}$ is a dibasic acid,its initial normality $N_{1} = M_{1} 	imes 	ext{basicity} = 0.2 	imes 2 = 0.4 \ N$.Final volume $V_{2} = 0.5 \ dm^{3}$.Using the dilution equation $N_{1}V_{1} = N_{2}V_{2}$:$0.4 \ N 	imes 2.5 	imes 10^{-3} \ dm^{3} = N_{2} 	imes 0.5 \ dm^{3}$.$N_{2} = \frac{0.4 	imes 2.5 	imes 10^{-3}}{0.5} = \frac{1 	imes 10^{-3}}{0.5} = 2 	imes 10^{-3} \ N = 0.002 \ N$.

$2.5 \ cm^{3}$ of $0.2 \ M \ H_{2}SO_{4}$ solution is diluted to $0.5 \ dm^{3}$. Find the normality of the diluted solution. (in $N$)

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