Question

Q12.A solution of NH4OH of density 2.04 g/ml is taken into consideration,
whose weight percentage is 20% .Calculate molarity, molality and mole fraction
for the following solution.​

Answer 1

Answer:

Correct option is

C

2.28 mol kg−1

Suppose the volume of solution is 1000 ml.

1.02 gm of acetic acid in 1 ml of soln.

1020 gm in 1000 ml of soln.

Number ofr moles =2⋅05M×1 lit=2.05 mol

Mass of solute =n×M⋅wt=2.05×60=123 gm

Mass of solvent =1020−123=897 gm

Molality =60123×8971000=2.28 mol kg−1