Question

Calculate the lowering of vapour pressure for a 3 moal aqueous solution of urea.​

Answer 1

Explanation:

Relative lowering vapour pressure

Molar weight of the water M=18g/mol

so formula for molarity=x×1000/(1-x)M

So

3=x×1000/(1-x)×18

⇒3(1-x)=x×1000×18

3-3x=x×18000

x=3/18003

So the relative lowering of vapour pressure=mole fraction of urea

ΔP/=x

ΔP=760×3/18003

ΔP=0.126torr