5.85 , g of NaCl is dissolved in 1 , L of pur | vedclass

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(C) $1$. Calculate the number of moles of $NaCl$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$. $2$. Cal

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$1.2 	imes 10^{20}$

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(C) $1$. Calculate the number of moles of $NaCl$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$. $2$. Calculate the molarity of the solution: $M = \frac{n}{V(L)} = \frac{0.1 \, mol}{1 \, L} = 0.1 \, M$. $3$. In $1 \, mL$ $(10^{-3} \, L)$,the number of moles of $NaCl$ is $0.1 	imes 10^{-3} = 10^{-4} \, mol$. $4$. Each $NaCl$ unit dissociates into $2$ ions ($Na^+$ and $Cl^-$). $5$. Total number of ions = $2 	imes 	ext{moles} 	imes N_A = 2 	imes 10^{-4} 	imes 6.022 	imes 10^{23} = 1.2044 	imes 10^{20}$ ions.

$5.85 \, g$ of $NaCl$ is dissolved in $1 \, L$ of pure water. Calculate the total number of ions in $1 \, mL$ of this solution.

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