0.43 g of a metal of valence 2 was dissolved | vedclass

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(D) The reaction of the metal $(M)$ with $H_2SO_4$ is: $M + H_2SO_4 ightarrow MSO_4 + H_2$. Total millimoles of $H_2SO_4$ taken = $50 \ mL 	imes 0.

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$24$

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(D) The reaction of the metal $(M)$ with $H_2SO_4$ is: $M + H_2SO_4 ightarrow MSO_4 + H_2$. Total millimoles of $H_2SO_4$ taken = $50 \ mL 	imes 0.5 \ M = 25 \ mmol$. Reaction with $NaOH$: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. Millimoles of $NaOH$ used = $14.2 \ mL 	imes 1 \ M = 14.2 \ mmol$. Since $1 \ mmol$ of $H_2SO_4$ reacts with $2 \ mmol$ of $NaOH$,the unreacted $H_2SO_4 = 14.2 / 2 = 7.1 \ mmol$. Millimoles of $H_2SO_4$ reacted with metal = $25 - 7.1 = 17.9 \ mmol = 0.0179 \ mol$. Since the metal has a valence of $2$,the stoichiometry is $1:1$,so moles of metal = $0.0179 \ mol$. Atomic weight of metal = $	ext{mass} / 	ext{moles} = 0.43 \ g / 0.0179 \ mol \approx 24.02 \ u$.

$0.43 \ g$ of a metal of valence $2$ was dissolved in $50 \ mL$ of $0.5 \ M \ H_2SO_4$ solution. The unreacted acid required $14.2 \ mL$ of $1 \ M \ NaOH$ solution for neutralization. The atomic weight of the metal is: (in $u$)

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