The degree of dissociation of `Ca(NO_(3))_(2)` in dilute aqueous solution containing 7.0g of salt per 100.0g of water at `100^(@)"C"` is 70 % . If the vapour pressure of water at `100^(@)"C"` is 760mm Hg, the vapour pressure of the solution is

Options

A. 748.3mm Hg
B. 1492.6mm Hg
C. 373.2mm Hg
D. 74.03mm Hg

Answer

A

Explanation

Moles of `Ca(NO_(3))_(2)=7/168` present in 100.0g of water
Since dissociation is 70%, the total number of particles
`=7/168xx0.7xx3=0.0875="n"`
Also, moles of solvent `=100/18=5055="N"`
Applying Raoults law
`("P"^(@)-"P"_("solution"))/"P"^(@)="n"/("n"+"N")`
Or `(760-"p")/760=0.087/(0.087+5055)`
Or ` 760-"p"=760[0.087/5.637]=760[0.0154]=11.704`
Or ` "p"=760-11.704=748.2"mm " " Hg"`