At `298 K` , the vapour pressure of water is `23.75 mm Hg`. Calculate the vapour pressure at the same temperature over `5%` aqueous solution of urea. `[CO(NH_(2))_(2)]`.

Explanation

`5%` aqueous solution of urea means that
Mass of solution = 100 g ,
Mass of solute, i.e, urea, `w_(2)=5g`
`therefore" Mass of solvent, i.e., water "w_(1)=100-5=95 g,`
Vapour pressure of pure water `(p^(@))` at 298 K = 23.75 mm
Vapour pressure of urea solution =, i.e, `p_(s)=?,`
Molar mass of water `(M_(1))="18 g mol"^(-1)`
Molar mass of urea `CO(NH_(2))_(2)`, i.e., `M_(2)=12+16+14xx2+2="60 g mol"^(-1)`
Applying Raoult's law,
`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))`
Substituting the values, we get
`(23.75 - p_(s))/(23.75)=(5)/(60)xx(18)/(95)`
`"or "23.75-p_(s)=(5)/(60)xx(18)/(95)xx23.75`
`"or "23.75-p_(s)=0.375`
`"or "p_(s)=23.375`
Thus, the vapour pressure of `5%` urea solution = 23.375 mm