A $6.50$ molal solution of $\mathrm{KOH}$ (aq.) has a density of $1.89 \mathrm{~g} \mathrm{~cm}^{-3}$. The molarity of the
solution is___________ $\mathrm{mol} \mathrm{dm}^{-3}$. (Round off to the Nearest Integer).
[Atomic masses: $\mathrm{K}: 39.0 \mathrm{u} ; \mathrm{O}: 16.0 \mathrm{u} ; \mathrm{H}: 1.0 \mathrm{u}$ ]
$6.5$ molal $\mathrm{KOH}=1000 \mathrm{gm}$ solvent has
$6.5$ moles $\mathrm{KOH}$
so $\mathrm{wt}$ of solute $=6.5 \times 56$
$=364 \mathrm{gm}$
wt of solution $=1000+364=1364$
Volume of solution $=\frac{1364}{1.89} \mathrm{~m} \ell$
Molarity $=\frac{\text { mole of solute }}{\mathrm{V}_{\text {solution }} \text { in Litre }}$
$=\frac{6.5 \times 1.89 \times 1000}{1364}$
$=9.00$
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