\(60 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}\) dissolved in \(1 \mathrm{dm}^{3}\) solvent, what is molality of solution ? (density \(=1 \cdot 25 \mathrm{~g} / \mathrm{cm}^{3}\) )

\(0.8 \mathrm{~m}\)
Given,
Volume \(=1 \mathrm{dm}^{3}=10^{3} \mathrm{~cm}^{3}\)
Density \(=1.25 \mathrm{~g} / \mathrm{cm}^{3}\)
Mass \(=60 \mathrm{~g}\)
We know that, molahty \((M)=\frac{\text { Moles }}{\text { Mass of solvents }}\)
\(\therefore\) Moles \(=\frac{\text { Given mass }}{\text { Molar mass }}\) \(=\frac{60 \mathrm{~g}}{60 \mathrm{~g} / \mathrm{mol}}\) \(=1 \mathrm{~mol}\)
Molality \((\mathrm{m})=\frac{\text { Moles }}{\text { Volume } \times \text { Density }}\)
\(\mathrm{m}=\frac{1}{1000 \mathrm{~cm}^{3} \times 1.25 \mathrm{~g} / \mathrm{cm}^{3}}\)
\(=\frac{1}{1.25 \mathrm{~g} \times 10^{3}}\)
\(\Rightarrow \frac{1}{1.25 \mathrm{~kg}}\)
\(=0.8 \mathrm{~m}\)