Question

2.4g of sodium carbonate is dissolved in water to give $245mL$ of the solution. Find the molarity of the solution (Molecular weight of $N{a_2}C{O_3} = 106.0$ )

Answer 1

Hint: Mass of the solute and the total volume of the solvent are involved in the process of finding the molarity of the solution. This depends on using the finding the number of moles of the solute in the given solution.

Complete step by step answer:
Here 245mL of the solution contains $2.4g$ of $N{a_2}C{O_3}$.
We know, for defining the molarity the mass of the solute present in 1000mL of the solution needs to be calculated.
Hence, $1000mL$ of the solution contains: $\dfrac{{2.4}}{{245}} \times 1000$ grams of $N{a_2}C{O_3}$ .

Therefore, $1000mL$ of the solution contains $9.796g$ of $N{a_2}C{O_3}$ .

$106g$ of solute is mixed in $1000mL$ of the solvent for $1mole$ of the solution.
(This is taken for $1mole$ as it is the molecular weight of sodium carbonate)

Therefore, $9.796g$ of a solute when mixed in $1000mL$ of the solvent, the number of moles of the solution will be: $\dfrac{{9.796}}{{106}}$
Hence the number of moles of the solute in the given solution is $0.0924\left( M \right)$.
Therefore, the molarity of the solution is $0.0924\left( M \right)$.

Therefore, in the given solution in which $2.4g$ of $N{a_2}C{O_3}$ is mixed with $245mL$ of water, then the resultant molarity of the mixture is $0.0924\left( M \right)$ . The molarity is much lower because of the high molecular weight as compared to that of the solute mixed in the solution.

Note: The molarity needs to be calculated based on the parameters of the number of moles of solute. The solute mass and volume both are important as the volume needs to be normalized to a particular level so that only variable character observed here is the mass which can be used to calculate the molarity.