Question

How do you calculate the molarity of each solution \[1.0mol\]$KCl$ in $750ml$ of solution and then $0.50mol$ $MgC{{l}_{2}}$ in $1.5l$ of solution?

Answer 1

Hint: Molarity is defined as the number of moles of solute per litre of the given solution. It is denoted with a symbol $M$ . A molarity is defined as one gram of the solute present in one litre of the solution. Then it is known as one molar of the solution.

Formula used-$M=\dfrac{n}{V\left( L \right)}$
Where, $M$ is the molarity of the solution
$n$ is the number of moles of solute
$V\left( L \right)$ is the volume of the solution in litres

Complete step-by-step answer:Let us see how to calculate molarity-
$M=\dfrac{n}{V\left( L \right)}$
Where, $M$ is the molarity of the solution
$n$ is the number of moles of solute
$V\left( L \right)$ is the volume of the solution in litres
Firstly, we will calculate the molarity of $1.0mol$ $KCl$ in $750ml$
$M=\dfrac{n}{V\left( L \right)}$
We will convert $750ml$ into litres
Hence the volume of the $KCl$ solution in litres- $0.750$
Now substituting the values in the given formula we get,
$M=\dfrac{1.0}{0.750}$
$M=1.33mol{{L}^{-1}}$
The molarity of $KCl$ solution is $1.33mol{{L}^{-1}}$
Similarly, we will calculate the molarity of $0.50mol$ $MgC{{l}_{2}}$ in $1.5L$
$M=\dfrac{n}{V\left( L \right)}$
Now, substituting the values in the given formula we get,
$M=\dfrac{0.50}{1.50}$
$M=0.33mol{{L}^{-1}}$
The molarity of the $MgC{{l}_{2}}$ solution is $0.33mol{{L}^{-1}}$

Additional information:The SI unit of the molarity is $mol{{L}^{-1}}$ . It is the most common unit that is used for the measurement of the concentration of the solution.
Molarity is used to calculate the amount of the solute and the volume of the solution.

Note: You should make sure that you don’t get confused between molarity and molality. The molarity is defined as the moles of the solute per volume of the solution in litres whereas molality is defined as the moles of the solute per mass of the solvent. It is denoted with the symbol $m$ .