What is the molarity of acid in vinegar conta | vedclass

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(B) $1$. The molar mass of acetic acid $(CH_3COOH)$ is calculated as: $(2 	imes 12) + (4 	imes 1) + (2 	imes 16) = 60 \ g/mol$. $2$. Given $4.0 \%$

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$0.68$

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(B) $1$. The molar mass of acetic acid $(CH_3COOH)$ is calculated as: $(2 	imes 12) + (4 	imes 1) + (2 	imes 16) = 60 \ g/mol$. $2$. Given $4.0 \%$ $(w/w)$ means $4.0 \ g$ of acetic acid in $100 \ g$ of vinegar solution. $3$. The volume of $100 \ g$ of solution is calculated using density: $V = \frac{mass}{density} = \frac{100 \ g}{1.02 \ g/mL} \approx 98.04 \ mL = 0.09804 \ L$. $4$. Moles of acetic acid = $\frac{mass}{molar \ mass} = \frac{4.0 \ g}{60 \ g/mol} \approx 0.06667 \ mol$. $5$. Molarity $(M) = \frac{moles \ of \ solute}{volume \ of \ solution \ in \ L} = \frac{0.06667 \ mol}{0.09804 \ L} \approx 0.68 \ M$.

What is the molarity of acid in vinegar containing $4.0 \%$ acetic acid by $(w/w)$ and having a density of $1.02 \ g/mL$ (in $M$)?

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