A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3} $. The molarity of solution is _ _ _ _ $mol.dm^{-3} $. (Round off to the nearest integer).
[Atomic masses: K = 39.0 u, O = 16.0 u, H = 1.0 u]
Solution
We have, $6.50 = \frac{{w \times 1000}}{{56 \times W}}$
$\therefore \frac{W}{w} = \frac{{1000}}{{6.50 \times 56}}$
Density $ = \frac{{w + W}}{V} = 1.89 \times 1000$ g/ltr
Molarity $ = \frac{w}{{56 \times V}} = \frac{{w \times 1.89 \times 1000}}{{56 \times (w + W)}} = \frac{{1.89 \times 1000}}{{56 \times \left( {1 + \frac{W}{w}} \right)}} = \frac{{1.89 \times 1000}}{{56 \times \left( {1 + \frac{{1000}}{{6.50 \times 56}}} \right)}} = 9$
[Note: $ltr = dm^3 = 1000 cm^3 $]
