A solution containing 200 mL of 0.5 M KCl i | vedclass

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(B) $1$. Calculate moles of $Cl^-$ from $KCl$: $n(Cl^-) = M 	imes V(L) = 0.5 	imes 0.2 = 0.1 \ mol$.$2$. Calculate moles of $Cl^-$ from $MgCl_2$: $1

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$0.15$

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(B) $1$. Calculate moles of $Cl^-$ from $KCl$: $n(Cl^-) = M 	imes V(L) = 0.5 	imes 0.2 = 0.1 \ mol$.$2$. Calculate moles of $Cl^-$ from $MgCl_2$: $19 \% \ w/v$ means $19 \ g$ in $100 \ mL$,so $50 \ mL$ contains $9.5 \ g$ of $MgCl_2$.Molar mass of $MgCl_2 = 24.3 + 2 	imes 35.5 = 95.3 \ g/mol \approx 95 \ g/mol$.Moles of $MgCl_2 = 9.5 / 95 = 0.1 \ mol$.Since $MgCl_2 ightarrow Mg^{2+} + 2Cl^-$,$n(Cl^-) = 2 	imes 0.1 = 0.2 \ mol$.$3$. Total moles of $Cl^- = 0.1 + 0.2 = 0.3 \ mol$.$4$. Total volume before dilution = $200 \ mL + 50 \ mL = 250 \ mL = 0.25 \ L$.$5$. Final volume after $8$ times dilution = $0.25 \ L 	imes 8 = 2 \ L$.$6$. Molarity of $Cl^-$ = $	ext{Total moles} / 	ext{Final volume} = 0.3 \ mol / 2 \ L = 0.15 \ M$.

$A$ solution containing $200 \ mL$ of $0.5 \ M$ $KCl$ is mixed with $50 \ mL$ of $19 \% \ w/v$ $MgCl_2$ and the resulting solution is diluted $8$ times. What is the molarity of the chloride ion in the final solution?

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