Question

Calculate the mass of $oxalic{\text{ }}acid$ ( ${H_2}{C_2}{O_{4.}}2{H_2}O$ ) needed to prepare $250ml$ , $1M$ aqueous solution of $oxalic{\text{ }}acid$ .

Answer 1

To find the answer of the question we will use the molarity formula and then will find the mass of the given compound which is $oxalic{\text{ }}acid$ . Also to answer the question, finding the molecular weight of $oxalic{\text{ }}acid$ is important.

Complete answer:
Molarity is defined as the number of moles of solutes present in one litre of solution. It is denoted by $M$ . Thus molarity is-
$M = \dfrac{{No.{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}solute}}{{{\text{Volume }}of{\text{ }}solution{\text{ }}in{\text{ Litre}}}}$
$M = \dfrac{n}{V}$
Where $n$ is the number of moles
The SI unit of molality is $mol/L$ .
Molarity also known as molar concentration is not often used in thermodynamics because the volume of most solutions slightly depends on the temperature due to thermal expansion. This problem is usually resolved by introducing temperature correction factors, or by using a temperature independent measure of concentration such as molality.
The molar concentration depends on the variation of the volume of the solution due to thermal expansion.
To calculate the number of moles, it is the ratio of the given mass of the substance or compound present in the sample to the weight of the substance by its molar mass.
Firstly we will calculate the molecular weight of $oxalic{\text{ }}acid$ . Atomic mass of the following are-
$C = 12$ , $H = 1$ and $O = 16$
Therefore the molecular weight = $1 \times 6 + 12 \times 6 + 16 \times 6$
$= 126g/mol$
Now putting the given values in the molarity formula
$M = \dfrac{n}{V}$
Substituting the given values,
$0.1M = \dfrac{n}{{0.025mol}}$
$n = \dfrac{m}{{molar{\text{ }}mass}}$
Substituting the given values,
$0.025mol = \dfrac{m}{{126}}$
$m = 3.15g$
Therefore the mass of $oxalic{\text{ }}acid$ is $3.15g$ .

Note:
Molarity and molality are two different ways of expressing concentration of solutions. Molarity is the number of moles of the solute in one litre of solution whereas molality is expressed as the number of moles in one kilogram of solvent.