The molar concentration of chloride ions in t | vedclass

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(C) For $NaCl$: $n(Cl^-) = M 	imes V = 3.0 \ M 	imes 0.300 \ L = 0.9 \ mol$. For $BaCl_{2}$: $BaCl_{2} ightarrow Ba^{2+} + 2Cl^-$. $n(Cl^-) = 2

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$5.0$

Vedclass · Answer

(C) For $NaCl$: $n(Cl^-) = M 	imes V = 3.0 \ M 	imes 0.300 \ L = 0.9 \ mol$. For $BaCl_{2}$: $BaCl_{2} ightarrow Ba^{2+} + 2Cl^-$. $n(Cl^-) = 2 	imes M 	imes V = 2 	imes 4.0 \ M 	imes 0.200 \ L = 1.6 \ mol$. Total moles of $Cl^- = 0.9 \ mol + 1.6 \ mol = 2.5 \ mol$. Total volume of solution = $300 \ mL + 200 \ mL = 500 \ mL = 0.5 \ L$. Molar concentration of $Cl^- = \frac{	ext{Total moles}}{	ext{Total volume}} = \frac{2.5 \ mol}{0.5 \ L} = 5.0 \ M$.

The molar concentration of chloride ions in the resulting solution of $300 \ mL$ of $3.0 \ M$ $NaCl$ and $200 \ mL$ of $4.0 \ M$ $BaCl_{2}$ will be (in $M$)

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