The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it

Explanation

1 molal solution contains 1 mol of the solute in 1 kg of water
` therefore ` Mole fraction of solute = `(1)/(1 + (1000//18)) = (1)/(1 + 55.5) = 0.0177`
Applying Raoult.s law, we have
`(p^0 -p_x)/(p^0) - x_2 " or " (12.3 -p_s)/(12.3) = 0.0177`
`p_s = 12.08 kPa`