A sample of sea water contains 47 ppm of ${\mathrm{NO}}_2^{-}$ ions. If the density of solution is $1.01g(mL{)}^{-}$ the approximate molarity of sea water is

 $\begin{array}{r}\mathrm{Mol} \mathrm{wt}. \mathrm{of} {\mathrm{NO}}_2^{-}=14+(2\times 16)+1\\ =47{\mathrm{gmol}}^{-}\end{array}$

${10}^6\mathrm{g}$ sea water contains ${\mathrm{NO}}_2^{-}\text{ions }=47\mathrm{g}=\frac{47}{47}=1\mathrm{mol}$

:. 1000 g sea water contains ${{\mathrm{NO}}_2}^{-}\text{ions }$

$\begin{array}{r}=\frac{1\times 1000}{{10}^6}=0.001\mathrm{mol}\\ =0.001\mathrm{M}\end{array}$

$\left[\because \right.$wt. of solution $=\text{ wt. of }{\mathrm{H}}_2\mathrm{O}=1000\mathrm{g}$

Volume of solution $\left.=\frac{1000}{\text{ density }1.01}\simeq 1000\mathrm{mL}.\right]$