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If the density of a $5 \mathrm{M}$ solution of $\mathrm{KCl}$ is $1.26 \mathrm{~g} \mathrm{~mL}^{-1}$, then find the molality of the solution. (molecular mass of $\mathrm{KCl}$ $=74.55 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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Verified Answer
The correct answer is:
$3.38 \mathrm{~m}$
5 molar solution means there are 5 moles of $\mathrm{KCl}$ salt in $1 \mathrm{~L}$.
Molecular weight of $\mathrm{KCl}=74.55 \mathrm{~g} \mathrm{~mol}^{-1}$
Hence, there are $5 \times 74.55 \mathrm{~g}$ in $1 \mathrm{~L}$ of water
$$
\text { Density }=\frac{\text { mass }}{\text { volume }}
$$
Mass of $1 \mathrm{~L}$ of solution $=1.26 \mathrm{~g} / \mathrm{mL} \times 1000 \mathrm{~mL}$
$$
=1260 \mathrm{~g}
$$
Mass of water in solution $=$ Mass of solution
- Mass of solute
$$
=1260-372.75=887.25 \mathrm{~g}
$$
Mass of water in solution $=0.887 \mathrm{~kg}$
$$
\text { Molality }=\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\frac{5}{0.887}
$$
Molality $=3.38 \mathrm{~m}$
Hence, molality of the solution is $3.38 \mathrm{~m}$.
Molecular weight of $\mathrm{KCl}=74.55 \mathrm{~g} \mathrm{~mol}^{-1}$
Hence, there are $5 \times 74.55 \mathrm{~g}$ in $1 \mathrm{~L}$ of water
$$
\text { Density }=\frac{\text { mass }}{\text { volume }}
$$
Mass of $1 \mathrm{~L}$ of solution $=1.26 \mathrm{~g} / \mathrm{mL} \times 1000 \mathrm{~mL}$
$$
=1260 \mathrm{~g}
$$
Mass of water in solution $=$ Mass of solution
- Mass of solute
$$
=1260-372.75=887.25 \mathrm{~g}
$$
Mass of water in solution $=0.887 \mathrm{~kg}$
$$
\text { Molality }=\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\frac{5}{0.887}
$$
Molality $=3.38 \mathrm{~m}$
Hence, molality of the solution is $3.38 \mathrm{~m}$.
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