Question

How to calculate the equivalent weight of $$N{a}_{2}C{O}_3$$?

Answer 1

Hint: The equivalent weight of an acid or base can be calculated by dividing the molecular weight of acid or base with the basicity or acidity. The equivalent weight of ionic salts can be determined by dividing the molecular weight of a salt with the valency of a salt, valency is given by the charge of the ions.

Complete answer: The chemical compound is sodium carbonate.
The molecular formula of sodium carbonate is $$N{a}_{2}C{O}_3$$.
It has two sodium atoms., one carbon atom and three oxygen atoms.
The molecular weight of Sodium carbonate $$\left(N{a}_{2}C{O}_3\right)$$ is given by
$$2\times 23+1\times 12+3\times 16=106$$
The molar mass of sodium is $$23$$
The molar mass of carbon is $$12$$
The molar mass of oxygen is $$16$$.
The given compound is a salt, it can ionize into ions.
The Sodium carbonate $$\left(N{a}_{2}C{O}_3\right)$$ ionizes to give two sodium ions and one carbonate ion.
$$N{a}_{2}C{O}_3\to 2N{a}^{+}+C{O_3}^{2-}$$
Thus, the charge on the ions is two, which can be called as valency of a salt.
The equivalent weight of a salt can be determined by dividing the molecular weight of a salt by valency of a salt.
Calculated molecular weight is $$106$$.
The valency of a salt is $$2$$.
The Equivalent weight will be $$\text{molecular weight}÷2$$
Thus, the equivalent weight is $$106÷2=53g{\left(mole\right)}^{-1}$$
Hence, the equivalent weight of Sodium carbonate $$\left(N{a}_{2}C{O}_3\right)$$ is $$53g{\left(eq\right)}^{-1}$$

Note:
The molecular weight is different from the equivalent weight, but for some compounds the molecular weight is the same as equivalent weight. Thus, the valency or acidity or basicity has to be calculated accurately. Thus, for salts the charge on the ions must be taken clearly.