Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50g urea `(NH_(2)CONH_(2))` is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Consider Raoult's law and formula for relative lowering in vapour pressure,
`(P_(A)^(0)-P_(s))/(P_(A)^(0))=(n_(B))/(n_(A))=(W_(B))/(M_(B))xx(M_(A))/(W_(A))`
Where, `(P_(A)^(0)-P_(s))/(P_(A)^(0))` is called relative lowering in vapour pressure.

Explanation

Step I : Calculation of vapour pressure of water for this solution.
AC Cording to Raoult's law,
`(P_(A)^(0))/(P_(A)^(0))=(n_(B))/(n_(A))=(W_(B)//M_(B))/(W_(A)//M_(A))=(W_(B))/(M_(B))xx(M_(A))/(W_(A))" "…(i)`
(Pure water) `P_(A)^(0)=23.8" mm",`
`W_(B)("urea")=50" g",W_(A)("water")=850" g"`
`M_(B)("urea")=60" g mol"^(-1),M_(A)("water")=180" g mol"^(-1)`
Placing the values in eq. (i)
`(P_(A)^(0)-P_(s))/(P_(A)^(0))=((50" g")xx(18" g mol"^(-1)))/((60" g mol"^(-1))xx(850" g"))=0.01762`
`(23.8-P_(s))/(238)=0.01762,23.8-P_(s)=0.4194`
`P_(s)=23.3806~~23.38" mm Hg"`
Step II : Calculation of relative lowering of vapour pressue
Relative lowering in vapour pressure `=(P_(A)^(0)-P_(s))/(P_(A)^(0))=((23.8-23.38)"mm")/((23.8" mm"))=0.0176`