Question

calculate the mass of Urea required in making 2.5 kg of 0.25 molal aqueous solution

Answer 1

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

Answer 2

Hey !!

Mass of required aqueous solution = 2.5 kg = 2500 g

0.25 molal aqueous solution of urea it means that 0.25 mole of urea is dissolved in 1000 grams of water.

Mass of water = 1000 g

Mass of urea = 0.25 mol

Molar mass of urea (H₂NCONH₂) = 4 × 1 + 2 × 14 + 1 × 12 + 1 × 16 = 60 g mol⁻¹

        NOTE :- Mass of urea = Number of moles of urea × Molar mass of urea  

     Mass of 0.25 moles of urea = 0.25 mol ×× 60 h mol⁻¹ = 15 g

Mass of solution = 1000 g + 15 g = 1015 g

1015 g of aqueous solution contains urea = 15 g

∴ 2500 g of aqueous solution will require urea  

         = 15 g / 1015 kg × 2500 g  

            = 36.95 g

HOPE IT HELPS YOU !!