Question

The lowering of vapour pressure of 0.45 molal aqueous solution at 100 degree Celcius is

Answer 1

Explanation:

Lowering of vapour pressure = x

A

1 molal aqueous solution means 1 mole of solute in 1000 ml of water

density of water = 1 g/ml

in one 1 ml of water = 1 g

in 1000 ml of water = 1000g

moles of water = 55.56 moles

so, x

A

=

n

A

+n

B

n

A

so, we get lowering of vapour presssure = 0.01768 atom or 13.43 mm of Hg