Search any question & find its solution
Question:
Answered & Verified by Expert
$60 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}$ dissolved in $1 \mathrm{dm}^{3}$ solvent, what is molality of solution ? (density $=1 \cdot 25 \mathrm{~g} / \mathrm{cm}^{3}$ )
Options:
Solution:
1727 Upvotes
Verified Answer
The correct answer is:
0.8m
$0.8 \mathrm{~m}$
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 $(\mathrm{m})=\frac{\text { Moles }}{\text { Volume } \times \text { Density }}$
$\mathrm{m}=\frac{1}{1000 \mathrm{~cm}^{3} \times 1.25 \mathrm{~g} / \mathrm{cm}^{3}}$
$=\frac{1}{1.25 \mathrm{~g} \times 10^{3}}$
$\Rightarrow \frac{1}{1.25 \mathrm{~kg}}$
$=0.8 \mathrm{~m}$
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 $(\mathrm{m})=\frac{\text { Moles }}{\text { Volume } \times \text { Density }}$
$\mathrm{m}=\frac{1}{1000 \mathrm{~cm}^{3} \times 1.25 \mathrm{~g} / \mathrm{cm}^{3}}$
$=\frac{1}{1.25 \mathrm{~g} \times 10^{3}}$
$\Rightarrow \frac{1}{1.25 \mathrm{~kg}}$
$=0.8 \mathrm{~m}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.