$60 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}$ dissolved in $1 \mathrm{dm}^{3}$ solvent, what is molality of…

$60 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}$ dissolved in $1 \mathrm{dm}^{3}$ solvent, what is molality of solution ? (density $=1 \cdot 25 \mathrm{~g} / \mathrm{cm}^{3}$ )
  1. 0.8m
  2. 0.4m
  3. 0.2m
  4. 0.6m

Solution

$0.8 \mathrm{~m}$ 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 $(\mathrm{m})=\frac{\text { Moles }}{\text { Volume } \times \text { Density }}$ $\mathrm{m}=\frac{1}{1000 \mathrm{~cm}^{3} \times 1.25 \mathrm{~g} / \mathrm{cm}^{3}}$ $=\frac{1}{1.25 \mathrm{~g} \times 10^{3}}$ $\Rightarrow \frac{1}{1.25 \mathrm{~kg}}$ $=0.8 \mathrm{~m}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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