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1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.
- a)a
- b)b
- c)c
- d)d
Correct answer is '2'. Can you explain this answer?
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1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the v...
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1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the v...
Given:
- Mass of oxalic acid (COOH)2 = 1.575 gm
- Volume of solution = 250 ml
- Volume of NaOH solution required for neutralization = 25 ml
- Concentration of NaOH solution = N/15
Calculations:
Step 1: Calculate the molarity of oxalic acid solution
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, we need to calculate the moles of oxalic acid (COOH)2. The molar mass of oxalic acid is:
C = 12.01 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of (COOH)2 = (12.01 * 2) + (16.00 * 2) + (1.01 * 4) = 90.04 g/mol
Moles of oxalic acid (COOH)2 = (mass of solute) / (molar mass) = 1.575 / 90.04 = 0.0175 mol
Now, we need to calculate the volume in liters:
Volume of solution = 250 ml = 0.25 L
Molarity of oxalic acid solution = 0.0175 mol / 0.25 L = 0.07 M
Step 2: Calculate the moles of NaOH
Moles of NaOH = (molarity of NaOH) * (volume of NaOH solution in liters)
Since the concentration of NaOH solution is given as N/15, we need to convert it to molarity:
1 N = 1 mol/L
N/15 = (1/15) mol/L
Moles of NaOH = (1/15) mol/L * 0.025 L = 0.00167 mol
Step 3: Calculate the moles of oxalic acid reacting with NaOH
From the balanced chemical equation, we can see that 1 mole of oxalic acid (COOH)2 reacts with 2 moles of NaOH:
(COOH)2 + 2NaOH -> Na2C2O4 + 2H2O
Therefore, moles of oxalic acid reacting with NaOH = 2 * 0.00167 = 0.00334 mol
Step 4: Calculate the moles of water in the oxalic acid
Moles of water = moles of oxalic acid - moles of oxalic acid reacting with NaOH
Moles of water = 0.0175 - 0.00334 = 0.01416 mol
Step 5: Calculate the value of x
From the formula of oxalic acid (COOH)2.xH2O, we know that 1 mole of oxalic acid is associated with x moles of water.
Therefore, x = (moles of water) / (moles of oxalic acid) = 0.01416 / 0.0175 = 0.81
The value of x is approximately 0.81, which corresponds to option 2.
- Mass of oxalic acid (COOH)2 = 1.575 gm
- Volume of solution = 250 ml
- Volume of NaOH solution required for neutralization = 25 ml
- Concentration of NaOH solution = N/15
Calculations:
Step 1: Calculate the molarity of oxalic acid solution
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, we need to calculate the moles of oxalic acid (COOH)2. The molar mass of oxalic acid is:
C = 12.01 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of (COOH)2 = (12.01 * 2) + (16.00 * 2) + (1.01 * 4) = 90.04 g/mol
Moles of oxalic acid (COOH)2 = (mass of solute) / (molar mass) = 1.575 / 90.04 = 0.0175 mol
Now, we need to calculate the volume in liters:
Volume of solution = 250 ml = 0.25 L
Molarity of oxalic acid solution = 0.0175 mol / 0.25 L = 0.07 M
Step 2: Calculate the moles of NaOH
Moles of NaOH = (molarity of NaOH) * (volume of NaOH solution in liters)
Since the concentration of NaOH solution is given as N/15, we need to convert it to molarity:
1 N = 1 mol/L
N/15 = (1/15) mol/L
Moles of NaOH = (1/15) mol/L * 0.025 L = 0.00167 mol
Step 3: Calculate the moles of oxalic acid reacting with NaOH
From the balanced chemical equation, we can see that 1 mole of oxalic acid (COOH)2 reacts with 2 moles of NaOH:
(COOH)2 + 2NaOH -> Na2C2O4 + 2H2O
Therefore, moles of oxalic acid reacting with NaOH = 2 * 0.00167 = 0.00334 mol
Step 4: Calculate the moles of water in the oxalic acid
Moles of water = moles of oxalic acid - moles of oxalic acid reacting with NaOH
Moles of water = 0.0175 - 0.00334 = 0.01416 mol
Step 5: Calculate the value of x
From the formula of oxalic acid (COOH)2.xH2O, we know that 1 mole of oxalic acid is associated with x moles of water.
Therefore, x = (moles of water) / (moles of oxalic acid) = 0.01416 / 0.0175 = 0.81
The value of x is approximately 0.81, which corresponds to option 2.
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1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer?
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1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. 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 1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer?.
1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. 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 1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer?.
Solutions for 1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.a)ab)bc)cd)dCorrect answer is '2'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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