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

Parth Shrivastava
34 Points
11 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
6 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
Aparna
15 Points
5 years ago
Given,
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
(2500-x)=200x/3
2500=203x/3
x=7500/203
x=36.946 g

aeysha
28 Points
5 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
=0.625
molar mass of urea(NH2CONH2)=14+2*1+12+16+14+2*1
=60g
mass of urea=0.625*60=37.5

Rishi Sharma