Question

A $6.50$ molal solution of $KOH\left(aq\right)$ has a density of $1.89{\text{g cm}}^{-3}$. The molarity of the solution is ${\text{mol dm}}^{-3}$. (Round off to the nearest integer).
$[\text{Atomic masses}:K=39.0\text{u},O=16.0\text{u},H=1.0\text{u}]$

Answer 1

By using the relation,
$m=\frac{1000\times M}{(1000\times d)-M\times M_{solute}}\text{Where, m=molality of the solution}=6.50\text{m}\text{M=molarity of the solution}\text{d=density}=1.89{\text{g cm}}^{-3}{M}_{solute}=\text{Molar mass of solute}=39+16+1=56\text{u}\text{Substituting the values, we get}6.5=\frac{1000\times M}{1890-M\times 56}\Rightarrow 12285-364M=1000M\Rightarrow 1364M=12285\Rightarrow M=9{\text{mol dm}}^{-3}$