What is the molarity of a $0.500$ molal aqueous solution of calcium nitrate that has a density of $1.045$ ${\mathrm{gmL}}^{-1}$

$\text{ Molarity }=0.5\mathrm{mol}/\mathrm{kg}$

$\text{ densily }=1.045\mathrm{gm}/\mathrm{L}=\frac{1.045\times {10}^{-3}\mathrm{kg}/{\mathrm{m}}^3}{{10}^{-3}}$

$\begin{array}{r}\mathit{1}\mathit{ }L=1000\mathrm{ml}\\ =1000{\mathrm{cm}}^3\end{array}$

$=1.045\mathrm{kg}/{\mathrm{m}}^3$

$\text{ Molarity }=\frac{\text{ moles }}{\text{ has }\left(\mathrm{kg}\right)}\Rightarrow \text{ Mass }\overline{\left(k\right)=}\frac{\text{ moles }}{\text{ Molarity }}=\frac{\mathrm{moles}}{0.5}$

$\text{ Volume }=\frac{\text{ mass }}{\text{ dansity }}=\frac{\text{ moles }}{0.5}\times \frac{1}{1.045}\left({\mathrm{m}}^3\right)$

$\text{ Molarity }=\frac{\text{ moles }}{\text{ volume }}$

$=\frac{\mathrm{moles}}{\frac{\text{ moles }}{0.5}\times \frac{1}{1.045}\times {10}^{-3} \mathrm{L}}$$=\frac{0.5\times 1.045}{{10}^3}$$=0.5225\times {10}^{-3}(\mathrm{mol}/\mathrm{L})$