Question

A $6.50$ molal solution of $KOH (aq)$ has a density of $1.89\, g\, cm ^{-3} .$ The molarity of the solution is______ $\quad mol dm ^{-3} .($ Round off to the Nearest Integer).
[Atomic masses: $K : 39.0 u ; O : 16.0 u ; H : 1.0 u ]$

Answer 1

$6.5$ molal $KOH =1000 gm$ solvent has $6.5$ moles $KOH$

so wt of solute $=6.5 \times 56$

$=364\,gm$

wt of solution $=1000+364=1364$

Volume of solution $=\frac{1364}{1.89}\,m \ell$

Molarity $=\frac{\text { mole of solute }}{ V _{\text {solution }} \text { in Litre }}$

$=\frac{6.5 \times 1.89 \times 1000}{1364}$

$=9.00$