(D) $1$. Calculate the mass of pure $H_2SO_4$ required: $n = M \times V(L) = 0.5 \ mol \ L^{-1} \times 0.5 \ L = 0.25 \ mol$. Mass of $H_2SO_4 = n \ti

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(D) $1$. Calculate the mass of pure $H_2SO_4$ required: $n = M \times V(L) = 0.5 \ mol \ L^{-1} \times 0.5 \ L = 0.25 \ mol$. Mass of $H_2SO_4 = n \times \text{molar mass} = 0.25 \ mol \times 98.079 \ g \ mol^{-1} = 24.51975 \ g \approx 24.52 \ g$. $2$. Calculate the mass of the $90\%$ solution required: Since the solution is $90\%$ pure,$\text{Mass of solution} = \frac{\text{Mass of solute}}{\text{Percentage}} = \frac{24.51975 \ g}{0.90} = 27.244 \ g \approx 27.24 \ g$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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What amount of conc. $H_2SO_4$ solution should be used to prepare $500 \ mL$ of $0.5 \ M \ H_2SO_4$ (in $g$)? (The concentration of $H_2SO_4$ solution being used is $90\%$ and molecular mass of $H_2SO_4 = 98.079 \ g \ mol^{-1}$)

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A
$22.06$
B
$24.52$
C
$11.03$
D
$27.24$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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What amount of conc. $H_2SO_4$ solution should be used to prepare $500 \ mL$ of $0.5 \ M \ H_2SO_4$ (in $g$)? (The concentration of $H_2SO_4$ solution being used is $90\%$ and molecular mass of $H_2SO_4 = 98.079 \ g \ mol^{-1}$)

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How many grams of $HCl$ are required to prepare $1500 \, mL$ of $0.1 \, N \, HCl$ aqueous solution?

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