Question

How do you calculate the number of grams of substance needed to make the following solution:
$50c{m^3}$ of $NaOH\left( {aq} \right)$, concentration $2mold{m^{ - 3}}$?

Answer 1

The answer to the question lies on the concept of molarity. It is a unit used to express the concentration of a solution in chemistry. A $1M$ solution is equal to $1mol/L$ of the solution.

Complete step by step answer:
A solution is composed of two parts of which one is the solute which is being dissolved and the other is the solvent in which the solute is dissolved. Upon dissolving the solute in solvent a solution is obtained which has a specific concentration.
Several concentration units are frequently used to express the concentration. The molarity, normality, molality are the common units of concentration.
In this question the molarity unit is used for expressing the concentration of the solution prepared. The solute in this question is $NaOH$ and the solvent in this question is water as the solution is aqueous $NaOH$.
The volume of the solution given is $50c{m^3}$ and the concentration of the aq. $NaOH$ is $2mold{m^{ - 3}}$. The molarity of a solution is defined as the number of moles of solute which is dissolved per liter of the solution.
So at first we need to calculate the number of moles of $NaOH$ for preparing the solution with given concentration and volume.
Thus the moles of $NaOH$=$50c{m^3} \times \dfrac{{2mol}}{{1d{m^3}}}$
$= 50c{m^3} \times \dfrac{{2mol}}{{1000c{m^3}}}$
$= 0.1mole.$
For determining the amount of substance we need the molar mass of that substance. The molar mass of $NaOH$ = atomic mass of $Na$ + atomic mass of $O$ + atomic mass of $H$
$= 23 + 16 + 1 = 40g/mol.$
Thus the amount of $NaOH$ required is = moles of $NaOH$ $\times$ molar mass of $NaOH$
$= 0.1mol \times \dfrac{{40g}}{{1mol}} = 0.4g$.

Hence, $0.4g$ of $NaOH$ is required for making the solution of volume $50c{m^3}$ and concentration $2mold{m^{ - 3}}$.

Note: The concentration is proportional to the volume of the solution made. As the volume of the solution increases the dilution of the solution increases thereby results in decrease in concentration of the solution.