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A 10 mg effervescent tablet contianing 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]
- a)16.8
- b)8.4
- c)0.84
- d)33.6
Correct answer is option 'B'. Can you explain this answer?
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A 10 mg effervescent tablet contianing sodium bicarbonate and oxalic a...
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A 10 mg effervescent tablet contianing sodium bicarbonate and oxalic a...
Solution:
Given:
Mass of effervescent tablet (m) = 10 mg
Volume of CO2 evolved (V) = 0.25 ml
Temperature (T) = 298.15 K
Pressure (p) = 1 bar
Molar volume of CO2 (Vm) = 25.0 L/mol
To find: Percentage of sodium bicarbonate in each tablet
Let's assume that the effervescent tablet contains x grams of sodium bicarbonate (NaHCO3).
Step 1: Calculate the number of moles of CO2 evolved.
We know that,
Volume of a gas (V) = Number of moles (n) * Molar volume (Vm)
n = V / Vm
Given,
V = 0.25 ml = 0.25 * 10^(-3) L
Vm = 25.0 L/mol
n = (0.25 * 10^(-3)) / 25.0 = 1 * 10^(-5) mol
Step 2: Calculate the molar mass of CO2.
The molar mass of CO2 can be calculated by adding the atomic masses of carbon (C) and two oxygen (O) atoms.
Molar mass of CO2 = (12.01 g/mol) + 2 * (16.00 g/mol) = 44.01 g/mol
Step 3: Calculate the mass of CO2 evolved.
Mass of CO2 (mCO2) = n * Molar mass of CO2
mCO2 = (1 * 10^(-5)) * 44.01 = 4.401 * 10^(-4) g
Step 4: Calculate the mass of sodium bicarbonate.
The reaction between sodium bicarbonate and oxalic acid can be represented as follows:
2 NaHCO3 + H2C2O4 -> Na2C2O4 + 2 CO2 + 2 H2O
From the balanced chemical equation, we can see that 2 moles of sodium bicarbonate react to produce 2 moles of CO2.
Therefore, the mass of sodium bicarbonate (mNaHCO3) can be calculated as follows:
mNaHCO3 = (mCO2 / 2) * (Molar mass of NaHCO3 / Molar mass of CO2)
mNaHCO3 = (4.401 * 10^(-4) / 2) * (84 / 44.01) = 4.401 * 10^(-4) * 84 / 88.02 = 4.2 * 10^(-4) g
Step 5: Calculate the percentage of sodium bicarbonate.
Percentage of sodium bicarbonate = (mNaHCO3 / m) * 100
Percentage of sodium bicarbonate = (4.2 * 10^(-4) / 10) * 100 = 4.2 * 10^(-3) %
Therefore, the percentage of sodium bicarbonate in each tablet is 0.0042%.
Hence, the correct answer is option 'B'.
Given:
Mass of effervescent tablet (m) = 10 mg
Volume of CO2 evolved (V) = 0.25 ml
Temperature (T) = 298.15 K
Pressure (p) = 1 bar
Molar volume of CO2 (Vm) = 25.0 L/mol
To find: Percentage of sodium bicarbonate in each tablet
Let's assume that the effervescent tablet contains x grams of sodium bicarbonate (NaHCO3).
Step 1: Calculate the number of moles of CO2 evolved.
We know that,
Volume of a gas (V) = Number of moles (n) * Molar volume (Vm)
n = V / Vm
Given,
V = 0.25 ml = 0.25 * 10^(-3) L
Vm = 25.0 L/mol
n = (0.25 * 10^(-3)) / 25.0 = 1 * 10^(-5) mol
Step 2: Calculate the molar mass of CO2.
The molar mass of CO2 can be calculated by adding the atomic masses of carbon (C) and two oxygen (O) atoms.
Molar mass of CO2 = (12.01 g/mol) + 2 * (16.00 g/mol) = 44.01 g/mol
Step 3: Calculate the mass of CO2 evolved.
Mass of CO2 (mCO2) = n * Molar mass of CO2
mCO2 = (1 * 10^(-5)) * 44.01 = 4.401 * 10^(-4) g
Step 4: Calculate the mass of sodium bicarbonate.
The reaction between sodium bicarbonate and oxalic acid can be represented as follows:
2 NaHCO3 + H2C2O4 -> Na2C2O4 + 2 CO2 + 2 H2O
From the balanced chemical equation, we can see that 2 moles of sodium bicarbonate react to produce 2 moles of CO2.
Therefore, the mass of sodium bicarbonate (mNaHCO3) can be calculated as follows:
mNaHCO3 = (mCO2 / 2) * (Molar mass of NaHCO3 / Molar mass of CO2)
mNaHCO3 = (4.401 * 10^(-4) / 2) * (84 / 44.01) = 4.401 * 10^(-4) * 84 / 88.02 = 4.2 * 10^(-4) g
Step 5: Calculate the percentage of sodium bicarbonate.
Percentage of sodium bicarbonate = (mNaHCO3 / m) * 100
Percentage of sodium bicarbonate = (4.2 * 10^(-4) / 10) * 100 = 4.2 * 10^(-3) %
Therefore, the percentage of sodium bicarbonate in each tablet is 0.0042%.
Hence, the correct answer is option 'B'.
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A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer?
Question Description
A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer?.
A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 mg effervescent tablet contianing 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]a)16.8b)8.4c)0.84d)33.6Correct answer is option 'B'. Can you explain this answer?.
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