Question

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

Answer 1

Answer: 9

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

so wt of solute $=6.5\times 56$

$=364gm$

wt of solution $=1000+364=1364$

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

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

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

$=9.00$