Question

What is the volume of ${\text{6}}{\text{.0M HN}}{{\text{O}}_{\text{3}}}$ solution if $0.837$ mole of ${\text{HN}}{{\text{O}}_{\text{3}}}$ are required in a particular reaction?

Answer 1

In the above question, we are provided with molarity and number of moles of ${\text{HN}}{{\text{O}}_{\text{3}}}$ is given and we have to find the volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution. For this, we can directly use molarity formula which is the ratio of number of moles of solute by volume of solution in litre.

Formula Used
${\text{Molarity (M) = }}\dfrac{{\text{n}}}{{\text{V}}}$
Where n= number of moles of solute
V= volume of solution in litre.

Complete step by step solution:
Molarity is the amount of a substance in a certain volume of solution. It can be defined as the moles of a solute per litre of a solution. It is also known as the molar concentration of a solution.
We know that, ${\text{Molarity (M) = }}\dfrac{{\text{n}}}{{\text{V}}}$
Rearranging the above equation, we get:
${\text{V = }}\dfrac{{\text{n}}}{{\text{M}}}$
In the above question, it is given that molarity of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is ${\text{6}}{\text{.0M}}$ and the number of moles is ${\text{0}}{\text{.837}}$ . So, M = ${\text{6}}{\text{.0M}}$ and n = ${\text{0}}{\text{.837}}$ .
So, by substituting the values, we get:
${\text{V = }}\dfrac{{\text{n}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.837}}}}{{\text{6}}}{\text{ = 0}}{\text{.139L}} \simeq {\text{0}}{\text{.14L}}$
$\therefore$ Volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ solution is ${\text{0}}{\text{.14L}}$ .

Note:
Sometimes, there is a confusion between molarity and molality. In such a case, remember, the difference between molarity and molality lies in the denominator part of the ratio. In molarity, we divide the number of moles by volume of the solution. And in molality, we divide the number of moles by mass of solvent in kg.
Solution=Solute + Solvent
So, in molality solute mass must be subtracted to get the correct result.
In molarity, the unit of the volume of the solution should be taken into consideration. It should be converted into litres in order to get the correct result.