If $20.0 \mathrm{~cm}^{3}$ of $1.0 \mathrm{M} \mathrm{CaCl}_{2}$ and $60.0 \mathrm{~cm}^{3}$ of $0.20 \mathrm{M} \mathrm{CaCl}_{2}$ are mixed, the molarity of the resultant solution is

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Aicharya · Accepted Answer

The molarity of the resultant solution will be
$$
M=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}}=\frac{(1.0 \mathrm{M})\left(20 \mathrm{~cm}^{3}ight)+(0.20 \mathrm{M})\left(60.0 \mathrm{~cm}^{3}ight)}{\left(20 \mathrm{~cm}^{3}+60 \mathrm{~cm}^{3}ight)}=\frac{20+12}{80} \mathrm{M}=0.40 \mathrm{M}
$$ ,

If $20.0 \mathrm{~cm}^{3}$ of $1.0 \mathrm{M} \mathrm{CaCl}_{2}$ and $60.0 \mathrm{~cm}^{3}$ of $0.20…

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