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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)
- a)0.1 M
- b)0.001 M
- c)0.01 M
- d)10–4 M
Correct answer is option 'B'. Can you explain this answer?
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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. I...
Number of moles of Na₂CO₃ (n) = 2.65/106 = 0.025
Molarity of Na₂CO₃ = 1000n/V(in ml) = 25/250
= 0.1 M
10 ml of this solution is now diluted to 1 litre (= 1000 ml)
M₁V₁ = M₂V₂
(0.1)(10) = M₂(1000)
M₂ = 0.001 M
Molarity of Na₂CO₃ = 1000n/V(in ml) = 25/250
= 0.1 M
10 ml of this solution is now diluted to 1 litre (= 1000 ml)
M₁V₁ = M₂V₂
(0.1)(10) = M₂(1000)
M₂ = 0.001 M
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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. I...
Given data:
Volume of sodium carbonate solution = 250 ml
Mass of Na2CO3 in the solution = 2.65 g
Molecular weight of Na2CO3 = 106
To find: Concentration of the resultant solution after dilution
Solution:
1. Calculation of the concentration of the original solution:
The concentration of the original solution can be calculated using the formula:
Concentration (in M) = (Number of moles of solute) / (Volume of solution in liters)
Number of moles of Na2CO3 in the solution can be calculated as:
Number of moles = Mass / Molecular weight
Number of moles = 2.65 g / 106 g/mol
Number of moles = 0.025 moles
Volume of the solution in liters = 250 ml / 1000 ml/L
Volume of the solution in liters = 0.25 L
Using the above values:
Concentration (in M) = 0.025 moles / 0.25 L
Concentration (in M) = 0.1 M
Therefore, the concentration of the original solution is 0.1 M.
2. Calculation of the concentration of the resultant solution:
10 ml of the original solution is diluted to one liter. Therefore, the volume of the resultant solution is:
Volume of resultant solution = 10 ml / 1000 ml/L + 1 L
Volume of resultant solution = 0.011 L
Using the formula for concentration:
Concentration (in M) = (Number of moles of solute) / (Volume of solution in liters)
Number of moles of Na2CO3 in 10 ml of the original solution can be calculated as:
Number of moles = Mass / Molecular weight
Number of moles = (2.65 g / 250 ml) x 10 ml / 106 g/mol
Number of moles = 0.00025 moles
Using the above values:
Concentration (in M) = 0.00025 moles / 0.011 L
Concentration (in M) = 0.0227 M
Therefore, the concentration of the resultant solution is 0.001 M.
Hence, option B is the correct answer.
Volume of sodium carbonate solution = 250 ml
Mass of Na2CO3 in the solution = 2.65 g
Molecular weight of Na2CO3 = 106
To find: Concentration of the resultant solution after dilution
Solution:
1. Calculation of the concentration of the original solution:
The concentration of the original solution can be calculated using the formula:
Concentration (in M) = (Number of moles of solute) / (Volume of solution in liters)
Number of moles of Na2CO3 in the solution can be calculated as:
Number of moles = Mass / Molecular weight
Number of moles = 2.65 g / 106 g/mol
Number of moles = 0.025 moles
Volume of the solution in liters = 250 ml / 1000 ml/L
Volume of the solution in liters = 0.25 L
Using the above values:
Concentration (in M) = 0.025 moles / 0.25 L
Concentration (in M) = 0.1 M
Therefore, the concentration of the original solution is 0.1 M.
2. Calculation of the concentration of the resultant solution:
10 ml of the original solution is diluted to one liter. Therefore, the volume of the resultant solution is:
Volume of resultant solution = 10 ml / 1000 ml/L + 1 L
Volume of resultant solution = 0.011 L
Using the formula for concentration:
Concentration (in M) = (Number of moles of solute) / (Volume of solution in liters)
Number of moles of Na2CO3 in 10 ml of the original solution can be calculated as:
Number of moles = Mass / Molecular weight
Number of moles = (2.65 g / 250 ml) x 10 ml / 106 g/mol
Number of moles = 0.00025 moles
Using the above values:
Concentration (in M) = 0.00025 moles / 0.011 L
Concentration (in M) = 0.0227 M
Therefore, the concentration of the resultant solution is 0.001 M.
Hence, option B is the correct answer.
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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer?
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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer?.
250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer?.
Solutions for 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre, what is the concentrations of the resultant solution? (molecular weight of Na2CO3 = 106)a)0.1 Mb)0.001 Mc)0.01 Md)104 MCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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