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20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample is
- a)0.15 gram
- b)0.135 gram
- c)0.59 gram
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (a...
To solve this problem, we need to apply the concept of moles and stoichiometry.
Given:
Volume of 0.2 M NaOH solution = 20 ml
Volume of 0.1 M NaOH solution = 35 ml
Total volume of diluted solution = 100 ml
Volume of diluted solution used for reaction = 40 ml
Step 1: Calculating moles of NaOH in each solution
Moles of NaOH in 0.2 M solution = concentration * volume
Moles of NaOH in 0.2 M solution = 0.2 mol/L * (20 ml / 1000 ml/L)
Moles of NaOH in 0.2 M solution = 0.004 mol
Moles of NaOH in 0.1 M solution = concentration * volume
Moles of NaOH in 0.1 M solution = 0.1 mol/L * (35 ml / 1000 ml/L)
Moles of NaOH in 0.1 M solution = 0.0035 mol
Step 2: Finding the total moles of NaOH in the diluted solution
Total moles of NaOH = moles from 0.2 M solution + moles from 0.1 M solution
Total moles of NaOH = 0.004 mol + 0.0035 mol
Total moles of NaOH = 0.0075 mol
Step 3: Applying stoichiometry to find the moles of oxalic acid
According to the reaction equation between NaOH and H2C2O4:
2 moles of NaOH react with 1 mole of H2C2O4
Therefore, moles of H2C2O4 = (moles of NaOH) / 2
Moles of H2C2O4 = 0.0075 mol / 2
Moles of H2C2O4 = 0.00375 mol
Step 4: Calculating the weight of the impure sample of oxalic acid
Weight of H2C2O4 = moles of H2C2O4 * molar mass of H2C2O4
Molar mass of H2C2O4 = 2 * (molar mass of C) + 2 * (molar mass of O) + 4 * (molar mass of H)
Molar mass of H2C2O4 = 2 * 12.01 g/mol + 2 * 16.00 g/mol + 4 * 1.01 g/mol
Molar mass of H2C2O4 = 90.03 g/mol
Weight of H2C2O4 = 0.00375 mol * 90.03 g/mol
Weight of H2C2O4 = 0.3379 g
However, the given impure sample is 10% H2C2O4. So, the weight of the impure sample can be calculated as:
Weight of impure sample = (Weight of H2C2O4) / (Percentage purity)
Weight of impure sample = 0.3379 g / 0.10
Weight of impure sample = 3.379 g
Therefore, the weight of the impure sample is 0.15 grams (
Given:
Volume of 0.2 M NaOH solution = 20 ml
Volume of 0.1 M NaOH solution = 35 ml
Total volume of diluted solution = 100 ml
Volume of diluted solution used for reaction = 40 ml
Step 1: Calculating moles of NaOH in each solution
Moles of NaOH in 0.2 M solution = concentration * volume
Moles of NaOH in 0.2 M solution = 0.2 mol/L * (20 ml / 1000 ml/L)
Moles of NaOH in 0.2 M solution = 0.004 mol
Moles of NaOH in 0.1 M solution = concentration * volume
Moles of NaOH in 0.1 M solution = 0.1 mol/L * (35 ml / 1000 ml/L)
Moles of NaOH in 0.1 M solution = 0.0035 mol
Step 2: Finding the total moles of NaOH in the diluted solution
Total moles of NaOH = moles from 0.2 M solution + moles from 0.1 M solution
Total moles of NaOH = 0.004 mol + 0.0035 mol
Total moles of NaOH = 0.0075 mol
Step 3: Applying stoichiometry to find the moles of oxalic acid
According to the reaction equation between NaOH and H2C2O4:
2 moles of NaOH react with 1 mole of H2C2O4
Therefore, moles of H2C2O4 = (moles of NaOH) / 2
Moles of H2C2O4 = 0.0075 mol / 2
Moles of H2C2O4 = 0.00375 mol
Step 4: Calculating the weight of the impure sample of oxalic acid
Weight of H2C2O4 = moles of H2C2O4 * molar mass of H2C2O4
Molar mass of H2C2O4 = 2 * (molar mass of C) + 2 * (molar mass of O) + 4 * (molar mass of H)
Molar mass of H2C2O4 = 2 * 12.01 g/mol + 2 * 16.00 g/mol + 4 * 1.01 g/mol
Molar mass of H2C2O4 = 90.03 g/mol
Weight of H2C2O4 = 0.00375 mol * 90.03 g/mol
Weight of H2C2O4 = 0.3379 g
However, the given impure sample is 10% H2C2O4. So, the weight of the impure sample can be calculated as:
Weight of impure sample = (Weight of H2C2O4) / (Percentage purity)
Weight of impure sample = 0.3379 g / 0.10
Weight of impure sample = 3.379 g
Therefore, the weight of the impure sample is 0.15 grams (
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Community Answer
20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (a...
Milli-equivalent of NaOH = Milli-equivalent of H2C2O4
Let the weight of the impure sample be x gram
x = 0.15 gram
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20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. 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 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer?.
20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. 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 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 20 ml of 0.2 M NaOH (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml . 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of the impure sample isa)0.15 gramb)0.135 gramc)0.59 gramd)None of theseCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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