Question

X moles of k2so4 to be dissolved in 12 mol water to lower its vapour pressure by 15 mm hg at a temprature at which vapour pressure of pure water is

Answer 1

Let P be the vapour pressure of pure solvent

P' be the vapour pressure of solution

n be the number of moles of solute

N be the number of moles of water (solvent)

We know,

P-P'P=nn+N 1050=nn+12 n+12 = 5n n = 12/4 =3

Hence 3 moles of K2SO4 should be dissolved.