A 6.50 molal solution of KOH (aq.)


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).



$6.5$ 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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now