If 98 , mg of H_2SO_4 is present in 100 , g o | vedclass

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(C) Step $1$: Calculate the moles of solute $(H_2SO_4)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{98 	imes 10^{-3} \, g}{98 \, g/mol} = 10^

Vedclass · Accepted Answer

$1.25 	imes 10^{-2} \, M$

Vedclass · Answer

(C) Step $1$: Calculate the moles of solute $(H_2SO_4)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{98 	imes 10^{-3} \, g}{98 \, g/mol} = 10^{-3} \, mol$. Step $2$: Calculate the volume of the solution: $V = \frac{	ext{mass of solution}}{	ext{density}} = \frac{100 \, g}{1.25 \, g/mL} = 80 \, mL = 0.08 \, L$. Step $3$: Calculate molarity $(M)$: $M = \frac{n}{V(L)} = \frac{10^{-3} \, mol}{0.08 \, L} = 0.0125 \, M = 1.25 	imes 10^{-2} \, M$.

If $98 \, mg$ of $H_2SO_4$ is present in $100 \, g$ of solution,what is the molarity of the solution? (Given: density of solution = $1.25 \, g/mL$)

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