What will be the molarity of Cl^{-} ions in M | vedclass

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(B) The dissociation reactions are as follows:$NaCl \rightarrow Na^{+} + Cl^{-}$$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$Moles of $Cl^{-}$ from $NaCl = 0

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

(B) The dissociation reactions are as follows:$NaCl ightarrow Na^{+} + Cl^{-}$$BaCl_2 ightarrow Ba^{2+} + 2Cl^{-}$Moles of $Cl^{-}$ from $NaCl = 0.2 \, mol 	imes 1 = 0.2 \, mol$Moles of $Cl^{-}$ from $BaCl_2 = 0.2 \, mol 	imes 2 = 0.4 \, mol$Total moles of $Cl^{-} = 0.2 + 0.4 = 0.6 \, mol$Volume of solution = $500 \, mL = 0.5 \, L$Molarity of $Cl^{-} = \frac{	ext{Total moles of } Cl^{-}}{	ext{Volume of solution in } L} = \frac{0.6 \, mol}{0.5 \, L} = 1.2 \, M$

What will be the molarity of $Cl^{-}$ ions in $M$ when $0.2 \, mol$ of $NaCl$ and $0.2 \, mol$ of $BaCl_2$ are dissolved in $500 \, mL$ of water?

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