Question

A 5% solution of sucrose (c12h22o11) is isotonic with 0.877% solution of urea. Nh2conh2 ) calculate the molecular mass of urea.[ans. : 59.99 g mol1]​

Answer 1

Answer:

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

hope it helps

Explanation:

Answer 2

Answer:

To find:

Mass of urea

Given

5% solution of sucrose (c12h22o11)

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  solution contains  = 15×2500/1000+15

                                                               = 36.946 g

                                                               = 37 g of urea (approx.)

Hence, mass of urea required = 37 g

#SPJ2