Question:
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 _________moldm $^{-3}$. (Round off to the Nearest Integer).
Solution:
(9)
$6.5$ molal $\mathrm{KOH}=1000 \mathrm{gm}$ solvent has
$6.5$ moles $\mathrm{KOH}$
so $\mathrm{w}$ t 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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