Molality of aqueous solution of urea is 3. The mass percentage of urea in the solution is-
a) 10.5%
b) 25.25%
c) 12.25%
d) 15.25%
PS... the answer is (d)
Please tell me the method
Answers
Molality of aqueous solution of Urea is 3. it means, 3 mole of urea present in 1kg of water solvent.
molecular weight of urea = 60 g/mol
so, weight of 3 mole of urea = 3mol × 60 g/mol = 180 g
weight of water (solvent ) = 1kg = 1000g
so, mass of solution = mass of urea + mass of water = 180 + 1000 = 1180g
now, mass percentage of urea in the solution = mass of urea/mass of solution × 100
= (180/1180) × 100
= 1800/118
= 15.25 %
hence, option (D) is correct choice.
Molality is a measure of concentration which tells us the amount of solute present (moles) per unit mass (Kg) of solvent.
In the given question, Molality is 3, which means there are 3 moles of urea present in 1 Kg of solvent.
In order to convert it into mass percent, we need to know how much urea is present in terms of its mass.
Since molecular formula of Urea is CH4N2O, its molar mass is
12 + (4 x 1) + (14 x 2) + 16 = 60 g/mol
Mass of 1 mole of urea = 60 g
Mass of 3 moles of urea = 3 x 60 = 180 grams
Now, we have
Mass of urea (solute) = 180 grams
Mass of solvent = 1 Kg = 1000 grams
Mass of solution = mass of solute + mass of solvent
= 180 g + 1000 g
= 1180 grams
Mass percentage of solute = (mass of solute/mass of solution) x 100
Putting numerical values
Mass percentage of urea = (180/1180) x 100
= 15.25 %