Question

Molality of aqueous solution of urea is 3.the mass percentage of urea in the solution is

Answer 1

we are given molality of urea as 3 which basically means that 3 moles of urea is present in 1000g /1kg of solvent.

as we know:

molality = mass of solute/ (molar mass of solute * mass of solvent in kg)

molar mass of urea = 60 (NH2CONH2)

by putting the values:

3=mass of urea/(60*1)

mass of urea = 180g

mass of solute (urea) =80g

mass of solvent=1000g

mass of solution= 1000+180=1180g

mass percentage= mass of urea*100/mass of solution

mass%= 180*100/1180

              1800/1180

               15.25%

i hope u get that right..............

Answer 2

Answer:

18%

Explanation:

Given :

Molality of aqueous solution (m)  = 3

The formula of Urea = NH2CoNH2

Let us assume that weight be W

and,

Let the molar mass be 60

Therefore,

m= (W/M) / 1 kg.

3 = W/M

W = 3*60 = 180g.

To find the mass %,

(180/1000) * 100

=18%

Hence, Mass percentage of Urea = 18%