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The density of a solution prepared by dissolving 120 g of urea in 1000 g of water is 1.15 g/ml. The molarity of this solution is:
  • a)
    0.50 M
  • b)
    1.78 M
  • c)
    1.02 M
  • d)
    2.05 M
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The density of a solution prepared by dissolving 120 g of urea in 1000...
d = M/V  1.15 gm/ml
 ⟹ 1.15 =  1120/V
V = 1120/1.15 = 973.9 ml
973.9 ml - 120/60 = 2 mole
1000ml - 2 × 1000/973.9 = 2.05m
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Most Upvoted Answer
The density of a solution prepared by dissolving 120 g of urea in 1000...
Given data:
Mass of urea = 120 g
Mass of water = 1000 g
Density of solution = 1.15 g/ml

We need to find the molarity of the solution.

To find the molarity of the solution, we need to first calculate the number of moles of urea present in the solution.

Number of moles of urea = Mass of urea / Molar mass of urea

The molar mass of urea (NH2)2CO is:
(2 x 14.01 g/mol) + (1 x 12.01 g/mol) + (1 x 16.00 g/mol) = 60.06 g/mol

Number of moles of urea = 120 g / 60.06 g/mol = 1.998 mol

The volume of the solution can be calculated using the density of the solution and the mass of the solution.

Volume of solution = Mass of solution / Density of solution

Mass of solution = Mass of urea + Mass of water = 120 g + 1000 g = 1120 g

Volume of solution = 1120 g / 1.15 g/ml = 973.91 ml

Now we can calculate the molarity of the solution.

Molarity = Number of moles of solute / Volume of solution in liters

Volume of solution in liters = 973.91 ml / 1000 ml/L = 0.97391 L

Molarity = 1.998 mol / 0.97391 L = 2.05 M

Therefore, the molarity of the solution is 2.05 M, which is option (d).
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The density of a solution prepared by dissolving 120 g of urea in 1000...
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The density of a solution prepared by dissolving 120 g of urea in 1000 g of water is 1.15 g/ml. The molarity of this solution is:a)0.50 Mb)1.78 Mc)1.02 Md)2.05 MCorrect answer is option 'D'. Can you explain this answer?
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