Some Basic Concepts of Chemistry 1 Question 30
50. Calculate the molality of $1.0 \mathrm{~L}$ solution of $93 % \mathrm{H}{2} \mathrm{SO}{4}$, (weight/volume). The density of the solution is $1.84 \mathrm{~g} / \mathrm{mL}$.
(1990, 1M)
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Solution:
- $93 % \mathrm{H}{2} \mathrm{SO}{4}$ solution weight by volume indicates that there is $93 \mathrm{~g} \mathrm{H}{2} \mathrm{SO}{4}$ in $100 \mathrm{~mL}$ of solution.
If we consider $100 \mathrm{~mL}$ solution, weight of solution $=184 \mathrm{~g}$
Weight of $\mathrm{H}_{2} \mathrm{O}$ in $100 \mathrm{~mL}$ solution $=184-93=91 \mathrm{~g}$
$\Rightarrow$ Molality $=\frac{\text { Moles of solute }}{\text { Weight of solvent }(\mathrm{g})} \times 1000$
$$ =\frac{93}{98} \times \frac{1000}{91}=10.42 $$
