Question

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

Answer 1

Given,
molality = 0.25 molal
means, 1000g of water (solvent ) contains 0.25 moles of Urea (solute)
molar mass of Urea = 14 + 2 + 12 + 16 + 14 + 2
= 16 + 28 + 16 = 60 g/mol
weight of Urea = mole of Urea × molar mass of Urea = 0.25 × 60 = 15g
now, mass of solution = mass of solvent (water) + mass of solute ( Urea)
= 1000g + 15g = 1015g

means, 1015g of solution contains 15g of Urea
so, mass of solute (Urea) in 2.5kg = 15 × 2.5 × 10³/1015
= 15 × 2500/1015 = 36.95 g ≈ 37g

hence, weight of Urea = 37g

Answer 2

Answer :

Given,

Molality = 0.25 m

Mass of solution = 2.5 Kg

∴ Value in grams = 2.5 × 1000 = 2500 g

Molar mass of urea $(NH_{2}CONH_{2})$, $M_{B}$ = 14 + 2 × 1 + 12 + 16 + 14 + 2 × 1 = 60 g mol−1

Let mass of urea $(NH_{2}CONH_{2})$ = $W_{B}$

Now,

Mass of solution = 2500 g

Mass of solute = 0.25 × 60 = 15 g

∴ Mass of solvent = (2500 - 15 g) = 2485 g

m = $\frac{W_{B}}{M_{B}}$ × $\frac{\text{1000}}{\text{Mass of solvent in g}}$

⇒ 0.25 = $\frac{W_{B}}{60} \times \frac{1000}{2485}$

⇒ $W_{B} = \frac{60\times0.25\times2485}{1000}$

⇒ $W_{B}$ = 15 × 2.485 = 37.275

∴ Mass of urea is 37.275 g