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For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required to
reach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) is
reach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) is
- a)5 x 10–4g/mL
- b)2 x 10–4g/mL
- c)5 x 10–5g/mL
- d)2 x 10–5g/mL
Correct answer is option 'D'. Can you explain this answer?
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For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M N...
To determine the concentration of CaCO3 in the 10 mL solution, we need to use the stoichiometry of the reaction between CaCO3 and Na2EDTA.
The balanced chemical equation for the reaction is:
CaCO3 + Na2EDTA -> CaEDTA + Na2CO3
From the stoichiometry of the reaction, we can see that 1 mole of CaCO3 reacts with 1 mole of Na2EDTA. Therefore, the number of moles of CaCO3 in the 10 mL solution is equal to the number of moles of Na2EDTA used.
The volume of Na2EDTA used is given as 2 mL. To calculate the number of moles of Na2EDTA used, we can use the formula:
moles = concentration * volume
The concentration of Na2EDTA is given as 0.001 M:
moles Na2EDTA = 0.001 M * 2 mL = 0.002 moles
Since the stoichiometry of the reaction is 1:1, the number of moles of CaCO3 in the 10 mL solution is also 0.002 moles.
To find the concentration of CaCO3, we can use the formula:
concentration = moles / volume
The volume of the solution is given as 10 mL:
concentration CaCO3 = 0.002 moles / 10 mL
However, we need to convert mL to liters to match the units of the concentration:
concentration CaCO3 = 0.002 moles / 0.01 L
Finally, to find the concentration in mol/L, we divide the number of moles by the volume in liters:
concentration CaCO3 = 0.002 moles / 0.01 L = 0.2 mol/L
Therefore, the concentration of CaCO3 in the 10 mL solution is 0.2 mol/L.
The balanced chemical equation for the reaction is:
CaCO3 + Na2EDTA -> CaEDTA + Na2CO3
From the stoichiometry of the reaction, we can see that 1 mole of CaCO3 reacts with 1 mole of Na2EDTA. Therefore, the number of moles of CaCO3 in the 10 mL solution is equal to the number of moles of Na2EDTA used.
The volume of Na2EDTA used is given as 2 mL. To calculate the number of moles of Na2EDTA used, we can use the formula:
moles = concentration * volume
The concentration of Na2EDTA is given as 0.001 M:
moles Na2EDTA = 0.001 M * 2 mL = 0.002 moles
Since the stoichiometry of the reaction is 1:1, the number of moles of CaCO3 in the 10 mL solution is also 0.002 moles.
To find the concentration of CaCO3, we can use the formula:
concentration = moles / volume
The volume of the solution is given as 10 mL:
concentration CaCO3 = 0.002 moles / 10 mL
However, we need to convert mL to liters to match the units of the concentration:
concentration CaCO3 = 0.002 moles / 0.01 L
Finally, to find the concentration in mol/L, we divide the number of moles by the volume in liters:
concentration CaCO3 = 0.002 moles / 0.01 L = 0.2 mol/L
Therefore, the concentration of CaCO3 in the 10 mL solution is 0.2 mol/L.
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For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer?
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For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer?.
For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required toreach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) isa)5 x 10–4g/mLb)2 x 10–4g/mLc)5 x 10–5g/mLd)2 x 10–5g/mLCorrect answer is option 'D'. Can you explain this answer?.
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