A is a solution containing 0.050 mol. dm of tetraoxosulphate (VI) acid. B is a solution of anhydrous trioxocarbonate (IV).

(a) Put A into the burette and titrate with 20cm\(^3\) or 25cm\(^3\) portions of B using methyl orange as an indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used

(b) From your results and the information provided, calculate the:

(i) Concentration of solution B in mol. dm\(^{-3}\)

(ii) mass of sodium tetraoxosulphate (VI) that would be formed in solution of 1dm\(^3\) of solution B were neutralized by solution A

(iii) volume of carbon (IV) oxide at s.t.p. that would be liberated in (b)(ii) above. The equation for the reaction is: N\(_2\)CO\(_{3(aq)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\)  Na\(_2\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\) [O = 16; Na = 23, S = 32; molar volume of gases of s.t.p. = 22.4dm\(^3\)