JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 3)
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]
33.6
0.84
8.4
16.8
Explanation
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 = $${1 \over {25 \times {{10}^3}}} \times 0.25$$ 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
= $${{84 \times {{10}^{ - 5}}} \over {10 \times {{10}^{ - 3}}}} \times 100$$ = 8.4 %
$$ \therefore $$ 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 = $${1 \over {25 \times {{10}^3}}} \times 0.25$$ 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
= $${{84 \times {{10}^{ - 5}}} \over {10 \times {{10}^{ - 3}}}} \times 100$$ = 8.4 %
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