What mass of ethylene glycol when mixed with 225 g H2O will

What mass of ethylene glycol, when mixed with 225 g H2O, will reduce the equilibrium vapor pressure of

H2O from 1.00 atm to 0.800 atm at 100

(P-Po)/Po = X(b) * Po , Po = 1, P = 0.8 ,

Xb = mol fraction of solute ,

(1-0.8)/1 = 1(Xb) ,

Xb = 0.2,

Xb = moles of glycol/(moles of glycol + moles of water),

moles of water = 225/18 = 12.5, hence

Xb = 0.2 = (m/m+12.5) ,

m= moles of glycol = 3.125 ,

moles = 3.125 = mass/(62.07) ,

mass of glycol = 193.96 gm