Question

Find out the molarity of $30g\,of\,CO{(N{O_3})_2}.6{H_2}O$ in $4.3L$ of solution.

Answer 1

Hint: In order to answer this question, to find the molarity of the given compound in the solution, first we will find the Molar mass of the given compound to calculate the Number of moles in the given compound. And finally we will find the molarity with the help of the number of moles that we found.

Complete answer:
Given that-
Mass of the given compound $ = 30g$
Quantity of solution $ = 4.3L$
So, now we will apply the formula to find the number of moles of the given compound:
$\therefore No.\,of\,Moles = \dfrac{{mass\,of\,the\,compound}}{{Molar\,Mass}}$
As we don’t know the molar mass of the given compound, so we will first find the molar mass of the given compound, $CO{(N{O_3})_2}.6{H_2}O$ :
$\because $ Molar mass of $CO{(N{O_3})_2}.6{H_2}O$
$ = 58.7 + 2(14 + 3 \times 16) + 6(2 + 16) = 290.7g.mo{l^ - }$
So, now we can find the number of moles-
$ \Rightarrow No.\,of\,Moles = \dfrac{{30}}{{290.7}} = 0.1032moles$
Now, we can find the Molarity with the help of no. of moles by applying the formula of Molarity:-
$
  \therefore Molarity = \dfrac{{No.\,of\,Moles}}{{quantity\,of\,solution}} \\
  = \dfrac{{0.1032}}{{4.3}} = 0.024M \\
 $
Hence, the molarity of the given compound in the respective solution is $0.024M$ .

Note:
One of the most popular measures used to quantify the concentration of a solution is molarity $(M)$ , which shows the number of moles of solute per litre of solution \[\left( {moles.Litr{e^{ - 1}}} \right)\] . Molarity is a unit of measurement that can be used to determine the volume of a solvent or the amount of a solute.