QJEEM19Ch9775: JEE Main 2019 Chemistry - Class 11 JEEM19Ch9775 Solution: A 10 mg effervescent tablet containing sodium bicarbonate and…

2NaHCO₃ + (COOH)₂ $$\to$$ (COONa)₂ + 2H₂O + 2CO₂

$$\therefore$$ 1 mole of CO₂ is produced by 1 mole of NaHCO₃.

Given, volume of CO₂ produced = 0.25 ml

25 L of CO₂ contains 1 mol

$$\therefore$$ 0.25 ml of CO₂ contains = $$\frac{1}{25\times {10}^3}\times 0.25$$ moles

= 10^-5 moles

$$\therefore$$ Moles of NaHCO₃ = 10^-5 moles

Mass of NaHCO₃ = 10^-5 $$\times$$ 84 g

$$\therefore$$ % of NaHCO₃ in a tablet

= $$\frac{84\times {10}^{-5}}{10\times {10}^{-3}}\times 100$$ = 8.4 %