Help with this The vapor pressure of water at 25 C is 238 to

-The vapor pressure of water at 25 °C is 23.8 torr. Determine g/mol) that must be added to 500.0 g of water to change the vapor pressure to 23 20 the mass of glucose (molar mass 180. .1 torr. A. 72 g B. 152 g C. 103 g 115 g E. 36 g

Solution

P_solution     = X_solvent * P0_H2O

23.1           = X_solvent * 23.8

X_solvent      = 23.1/23.8   = 0.97

mole fraction of water Xsolvent   = 0.97

n H2O/nH2O + n glucose   = 0.97

no of moles of H2O (n H2O ) = W/G.M.Wt

                                                = 500/18   = 27.8

n H2O/nH2O + n glucose   = 0.97

27.8/27.8 + n glucose      = 0.97

27.8    = (27.8 + n glucose)*0.97

n glucose   = 0.85 moles

mass of glucose = no of moles * gram molar mass

                           = 0.85*180    = 152g

B. 152 >>>>answer