Question

15 gram urea is mixed with water to prepare 200 ml of its aqua solution find out active mass of urea​

Answer 1

Answer:

Solution is in the attachment

Answer 2

Explanation:

Moles of urea =15g/60g/mol

=>1/4mol

=>0.25mol

Active mass=moles of solute/volume of solution in litre

1000ml=1litre

200ml=?

200/1000=>2/10=>0.2=>1/5 liter

=>0.25/0.2

=>1/4/1/5

=>1/4x5/1

=>1.25mol/litre

Active mass of urea =1.25mol/litre