A sodium salt gives no precipitate with $MgSO X_{4}$ solution in the cold, but a white precipitate (assume $MgCO X_{3}$ ) and a colourless gas appear on boiling. A $0.84 g$ sample of the salt is boiled with excess $MgSO X_{4}$ solution. Using Na 23, H 1, C 12, O 16 and molar volume $22.4 L mo l^{- 1}$ at STP, the volume of gas liberated is

Answer: A

Normal carbonate would precipitate in the cold, so the salt is

NaHCO X_{3}

(M = 84), and

2 HCO X_{3} X^{-} + Mg X^{2 +} MgCO X_{3} + CO X_{2} + H X_{2} O

. Moles of salt

= 0.84/84 = 0.0100

, giving

0.00500 mol

CO X_{2}

. Volume

= 0.00500 \times 22400 = 112 mL