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What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?
- a)37 g
- b)25 g
- c)125 g
- d)27.5 g
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What is the mass of urea required for making 2.5 kg of 0.25 molal aque...
Mass of solvent = 1000 g
Molar mass of urea (NH2CONH2) = 60gmol−1
0.25 mole of urea = 0.25 × 60 = 15g
Total mass of solution = 100 + 15 = 1.015kg
1.015 kg of solution contain urea = 15g
2.5 kg of solution =
Molar mass of urea (NH2CONH2) = 60gmol−1
0.25 mole of urea = 0.25 × 60 = 15g
Total mass of solution = 100 + 15 = 1.015kg
1.015 kg of solution contain urea = 15g
2.5 kg of solution =
Most Upvoted Answer
What is the mass of urea required for making 2.5 kg of 0.25 molal aque...
To determine the mass of urea required for making a 0.25 molal aqueous solution, we need to use the formula:
Molality (m) = (moles of solute) / (mass of solvent in kg)
The given molality is 0.25 mol/kg, and we want to make 2.5 kg of the solution.
Let's assume the mass of urea required is 'x' grams.
Now, we know that the molar mass of urea (CO(NH2)2) is 60 g/mol.
To calculate the moles of solute, we can use the formula:
Moles = Mass / Molar mass
Moles of urea = x / 60
Next, we need to calculate the mass of solvent (water) in kg. Since we want to make a 2.5 kg solution, the mass of water will be:
Mass of water = 2.5 kg - x (mass of urea)
Now, let's substitute these values into the formula for molality:
0.25 = (x / 60) / (2.5 - x)
To simplify this equation, we can cross multiply:
0.25 * (2.5 - x) = x / 60
0.625 - 0.25x = x / 60
Now, let's solve for x:
0.625 = 0.25x + x / 60
Multiplying through by 60:
37.5 = 15x + x
Combining like terms:
37.5 = 16x
Dividing by 16:
x = 37.5 / 16
x ≈ 2.34 g
Since the question asks for the mass in grams, we can round off to 2 decimal places and get:
x ≈ 2.34 g
Therefore, the mass of urea required to make 2.5 kg of a 0.25 molal aqueous solution is approximately 2.34 grams, which is closest to option A (37 g).
Molality (m) = (moles of solute) / (mass of solvent in kg)
The given molality is 0.25 mol/kg, and we want to make 2.5 kg of the solution.
Let's assume the mass of urea required is 'x' grams.
Now, we know that the molar mass of urea (CO(NH2)2) is 60 g/mol.
To calculate the moles of solute, we can use the formula:
Moles = Mass / Molar mass
Moles of urea = x / 60
Next, we need to calculate the mass of solvent (water) in kg. Since we want to make a 2.5 kg solution, the mass of water will be:
Mass of water = 2.5 kg - x (mass of urea)
Now, let's substitute these values into the formula for molality:
0.25 = (x / 60) / (2.5 - x)
To simplify this equation, we can cross multiply:
0.25 * (2.5 - x) = x / 60
0.625 - 0.25x = x / 60
Now, let's solve for x:
0.625 = 0.25x + x / 60
Multiplying through by 60:
37.5 = 15x + x
Combining like terms:
37.5 = 16x
Dividing by 16:
x = 37.5 / 16
x ≈ 2.34 g
Since the question asks for the mass in grams, we can round off to 2 decimal places and get:
x ≈ 2.34 g
Therefore, the mass of urea required to make 2.5 kg of a 0.25 molal aqueous solution is approximately 2.34 grams, which is closest to option A (37 g).
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What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?a)37 gb)25 gc)125 gd)27.5 gCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?a)37 gb)25 gc)125 gd)27.5 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?a)37 gb)25 gc)125 gd)27.5 gCorrect answer is option 'A'. Can you explain this answer?.
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