An aqueous solution is prepared by dissolving pure crystals of Mohr’s salt $FeS{O}_4{\left(N{H}_4\right)}_{2}S{O}_4.6H_{2}O$  in water. Density of the above solution is $1.2g/mL$ and the solution contains $11.833\%FeS{O}_4{\left(N{H}_4\right)}_{2}S{O}_4$  by weight. Then find the mass of the salt dissolved in  $400mL$ volume of solution.
(Molar masses: $Fe=56,S=32,N=14,H=1,O=16$ )
 

Since molarity is defined as moles of solute per litre of solution, it is always recommended to consider  $1.0L$ of solution to determine molarity when the density is given.
Therefore, mass of one litre of the above solution
   $$\begin{gathered} =volume\times density \\ =1000mL\times 1.2g/mL=1200g \end{gathered}$$
      Mass of  $FeS{O}_4{\left(N{H}_4\right)}_{2}S{O}_4$ present in  $1.0L$ of above solution
 $\text{=\hspace{0.17em}Mass\hspace{0.17em}of\hspace{0.17em}}1.0L\text{\hspace{0.17em}solutio}n\times \frac{30}{100}=1200\times \frac{11.833}{100}=142g$
    Mass of $FeS{O}_4{\left(N{H}_4\right)}_{2}S{O}_4$  present in  $1.0L$ solution   
  $=\frac{142}{284}=0.5$
 $\Rightarrow Molarity\left(M\right)=0.5M$
   Moles of anhydrous salt in $400mL$  solution
 $=0.5\times 0.40$
 $=0.2$
Moles of hydrated salt dissolved = Moles of the anhydrous salt in solution
Moles of hydrated salt dissolved for$400mL$   of solution = 0.2
     Mass of hydrated salt dissolved  $=0.2\times 392=78.4g$