Equal moles of ${\mathrm{H}}_2\mathrm{O}$ and a solute (of negligible molar mass) are present in a solution. Hence, molarity of solution is :

 

Mass of ${\mathrm{H}}_2\mathrm{O}=18\times 0.5=9\mathrm{g}$

Volume of ${\mathrm{H}}_2\mathrm{O}=9\mathrm{mL}\left(\rho =1\mathrm{g}/{\mathrm{cm}}^3\right)$

$\text{ Molarity }=\frac{\text{ Moles of solute }}{\text{ Volume of solution(iitres) }}=\frac{0.5}{\frac{9}{1000}}=55.5\mathrm{mol}/\text{ litres }$