Question

determine the lowering in vapour pressure if 30 gram urea is added in 72 gram water and the water and the vapour pressure of pure water is 200 tor​

Answer 1

Answer:

Mass of urea =5 g

Mass of water =100−5=95 g

Number of moles of urea =

60

5

=0.083

Number of moles of water =

18

95

=5.278

∴ Total number of moles =5.278+0.083=5.361

Mole fraction of solvent =

5.361

5.278

P

s

= Mole fraction of solvent ×P

o

=

5.361

5.278

×23.5

= 23.14 mm Hg ≈ 23 mm Hg