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Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of
0.25 molal aqueous solution.


5 Answers

Parth Shrivastava
34 Points
8 years ago

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol−1

0.25 molar aqueous solution of urea means: 

1000 g of water contains 0.25 mol = (0.25 × 60)g of urea = 15 g of urea i.e, 

(1000 + 15) g of solution contains 15 g of urea 

15×2500/1000+15 g


Therefore, 2.5 kg (2500 g) of solution contains  = 15×2500/1000+15 = 36.946 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g

Tejaswani Sharda Prasoon
11 Points
3 years ago
Let mass of urea = w gram Molar mass of urea = 60 gramMoles of urea = w/60 Weight of urea in kg= w/1000Molality =moles of solute/ mass of solvent in kg0.25=(w/60)/{2.5- (w/1000)}w=36.9456 gram
15 Points
2 years ago
Molality(m) = 0.25 mol kg-1
mass of the solution = 2.5 kg = 2500 g
To find,
mass of urea required = ?
[Molality = given mass of solute/molar mass × 1000/mass of the solvent (in g)]
molar mass=(14+2+12+16+14+2) =60 g mol-1 
Let x be the mass of urea required.
mass of the solvent = (2500-x)
substituting the values,
0.25=x/60 ×1000/(2500-x)
(2500-x)=x/60 ×1000/0.25
x=36.946 g
28 Points
2 years ago
molality=number of moles of solute/mass of solvent in kg
0.25=moles of solute/2.5
moles of solute=0.25*2.5
molar mass of urea(NH2CONH2)=14+2*1+12+16+14+2*1
mass of urea=0.625*60=37.5
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Hello students,
The solution of the above problem is in the attached file.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.

645-1588_WhatsApp Image 2020-06-05 at 1.35.07 AM.jpeg

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