JEE Main Chemistry Class 11 Solutions Part 1 Questions

According to Raoult's Law
‌

P∘−Ps

Ps∘

=‌

WB×MA

MB×WA

. . . (i)
Here P∘= Vapour pressure of pure solvent,
Ps= Vapour pressure of solution
WB= Mass of solute, WA= Mass of solvent
MB= Molar mass of solute, MA= Molar Mass of solvent
Vapour pressure of pure water at 100∘C (by assumption = 760 torr )
By substituting values in equation (i) we get,
‌

760−Ps

Ps∘

=‌

18×18

180×178.2

. . . (ii)
On solving (ii) we get
Ps=752.4 torr