First, we need to write a balanced equation for the chemical reaction:

 H\(_2\)SO\(_4\)  +  Na\(_2\)CO\(_3\) →  Na\(_2\)SO\(_4\)  + H\(_2\)O  +  CO\(_2\)

Stoichiometrically, N\(_A\) = 1 , N\(_B\) = 1

Given:

V\(_B\) = 20cm\(^3\), V\(_A\) = 25cm\(^3\), C\(_A\) = ?

C\(_B\) =  0.53 g is dissolved in 100 cm\(^3\) of water

  then  Xg will be dissolved in 1000 cm\(^3\) of water (i.e 1dm\(^3\) = 1 Litre of water)

  Xg = \(\frac{{0.53}\times{1000}}{100}\)

  Xg = 5.3gdm\(^{-3}\) → This is the concentration of B in  gdm\(^{-3}\). Hence we must have to convert it to moldm\(^{-3}\) before calculating the concentration of the acid.

To convert to molar concentration, we use the relationship;

Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration (gdm^{-3}}{Molar mass}\)

But Molar mass of Na\(_2\)CO\(_3\) = (23 x 2) + 12 + (16 x 3) = 106g/mol

Molar concentration (moldm\(^{-3}\)) = \(\frac{5.3}{106}\)

 C\(_B\) = 0.05 moldm\(^{-3}\)

Using \(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)

\(\frac{C_A}{V_A}{N_B}\)= \(\frac{C_B}{V_B}{N_A}\)

C\(_A\) = \(\frac{{C_B}{V_B}{N_A}}{{V_A}{N_B}}\)

C\(_A\) = \(\frac{{0.05}\times{20}\times{1}}{{25}\times{1}}\)

C\(_A\) = 0.04 moldm\(^{-3}\)