Calculate the molality of a solution containing $20.7g$ of potassium carbonate dissolved in $500ml$ of solution (assume density of solution $ = 1g{\text{ m}}{{\text{L}}^{ - 1}}$

Hint: Composition of a solution can be expressed on the basis of its concentration. The terms dilute and concentrated provide a vague idea about the concentration of the solution and thus a quantitative representation is needed for expression of the concentration. The concentration of the solution depends on the amount of solute and solvent that are present in the solution.

Complete step by step answer:
For the quantitative representation of the concentration of the solution there are several ways that can be used, molality is one of those criteria.
Molality is represented by $m$ and can be calculated by the following formula
$Molality(m) = \dfrac{{{{Moles\; of \;solute}}}}{{{{Mass \;of \;solvent\; in\; kg}}}}$
The mass of potassium carbonate as given in the question is $20.7g$ . So the molar mass of the compound will be $138gmo{l^{ - 1}}$ since the empirical formula of the compound is ${K_2}C{O_3}$
From the above data we can calculate the number of moles of the compound as
$Moles = \dfrac{{mass}}{{molar\;mass}}$
$ \Rightarrow moles = \dfrac{{20.7}}{{138}} = 0.15mo{l^{ - 1}}$
Since the volume of the solution is given as $500ml$ we can calculate the mass of the solution as,
$mass{{\; of\; solution = 500mL}} \times {{1gm}}{{{L}}^{ - 1}} = 500g$
$Amount{\text{ of water = 500 - 20}}{\text{.7 = 479}}{\text{.3g}}$
So the molality = $\dfrac{{Moles\;{{ of \;solute}}}}{{Mass\;{{ of \;solvent\; in\; grams}}}} \times 1000$
$molality = \dfrac{{0.15mol}}{{479.3g}} \times 1000$

So, the Molality = $0.013$.

Note: There are other representations of the concentration of solution than molality.
While molality is represented by $m$ another representation, Molarity is represented by $M$ .
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as in the caseṣ of molality. It can be represented as
${{Molarity (M) = }}\dfrac{{{{Moles\; of \;solute}}}}{{{{Volume \;of \;solution \;in\; litre}}}}$