Raoult’s Law
Your Turn: Assume the solution containing 62.1 g (1 mol) of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg)
Solution (from before Xglycol=1/(1+13.9)=0.0672)
Xglycol = 0.0672 and so Xwater = ?
Because Xglycol + Xwater = 1
Xwater = 1.000 - 0.0672 = 0.9328
Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)