JEE Main 2019 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2019 (Online) 11th January Morning Slot
Question
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol–1]
[Molar mass of NaHCO3 = 84 g mol–1]
Choose an option
Show full solutionCorrect option: C
Correct answer
C8.4
Step-by-step explanation
2NaHCO3
+ (COOH)2 (COONa)2 + 2H2O + 2CO2
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
0.25 ml of CO2 contains = moles
= 10-5 moles
Moles of NaHCO3 = 10-5 moles
Mass of NaHCO3 = 10-5 84 g
% of NaHCO3 in a tablet
= = 8.4 %
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
0.25 ml of CO2 contains = moles
= 10-5 moles
Moles of NaHCO3 = 10-5 moles
Mass of NaHCO3 = 10-5 84 g
% of NaHCO3 in a tablet
= = 8.4 %
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