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  • Help me please, The density of a solution prepared by dissolving 120 g of urea ( mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL.The molarity of this solution is :

The density of a solution prepared by dissolving 120 g of urea ( mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL.

The molarity of this solution is :

  • Option 1)

    0.50 M

  • Option 2)

    1.78 M

  • Option 3)

    1.02 M

  • Option 4)

    2.05 M

 

Answers (4)

best_answer

Mass of solute taken = 120 g

Molecular mass of solute  = 60 u

Mass of solvent  = 1000 g

Density of solution  =1.15 g/mL

Total mass of solution  = 1000 +120 + =1120 g

V\! olume\: o\! f \: solution = \frac {Mass}{Density }= \frac{1120}{1.15}mL

Molarity = \frac {M\! ass \, o\! f \, solute/Molecular\, mass\, of \, solute }{V\! olume\, of \, solution }\times 1000

= \frac {120/60}{1120/1.15}\times 1000

=\frac{2\times 1000\times 1.15}{1120}= 2.05M


Option 1)

0.50 M

Option 2)

1.78 M

Option 3)

1.02 M

Option 4)

2.05 M

Posted by

solutionqc

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Posted by

vishal dange

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The idea here is that a solution's molarity tells you the number of moles of solute present in 1 L of solution.

So in order to calculate a solution's molarity, you essentially need to know the number of moles of solute present in exactly 1 L=103 mL of solution.

Start by using the molar mass of urea to calculate the number of moles present in your sample

120g*1 mole urea/60g=2 moles urea

Now, you know that your solution contains 120 g of urea, the solute, and 1000 g of water, the solvent. This implies that the total mass of the solution

mass solution = mass solute + mass solvent

will be equal to

mass solution=120 g + 1000 g=1120 g

You also know that this solution has a density . of 1.15 g mL−1, which means that every 1 mL of solution has a mass of 1.15 g.

Use the density of the solution to calculate its volume

1120g*1 mL/1.15g=973.9 mL

Now, your goal is to figure out the number of moles of solute present in 103 mL of solution, so use the known composition of the solution as a conversion factor to get

103mL solution*2 moles urea/973.9mL solution=2.0536 moles urea

You can thus say that the molarity of the solution is equal to

molarity = 2.05 mol L−1

Posted by

Amit Kumar

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