Grade 12Organic Chemistry

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

K RAMA CHANDRA

13 Years agoGrade 12

5 Answers

Parth Shrivastava

13 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

8 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

7 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

7 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

6 Years 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.
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