Calculate the increase in the boiling point,ΔT, for a 0.86 m solution of sucrose (dissolves as a molecule) in water.
Tb = .512 (water)ΔT = m Kf (solvent) ΔT = (.86 m) ( .512 C/m)
Sugar is dissolved in hot water at 45oC until the mole fraction of water decreases to 0.870. The vapor pressure of pure water at this temperature is 72 torr. What is the vapor pressure (in torr) of water for this solution?
44.1 g of the nonvolatile solute urea (M.W. = 60.06 g/mol) is dissolved in 261 g of water at 60oC. The vapor pressure of pure water at this temperature is 149 torr. What is the equilibrium vapor pressure (in torr) of water for this solution?
Nonvolatie solute urea = 44.1 g Nonvolatie solute urea (molar mass) = 60.6 (g/mol) Water = 261g44.1 g / 60.6 (g/mol) = .72607 mol Nonvolatie 261 g/ 18.016 (g/mol) = 14.487 mol Water x = 14.487 /(.72607 + 14.487) = 0.9523Psolution = xsolvent PsolventPsolution = 0.9523 x 149 torr = 142 torr
Which 0.06 m aqueous solution (in water) should have the greatest increase in boiling point?
Large amount of heat which is needed to break the strong ionic bond.
Calculate the freezing point for a 3.93 m aqueous solution of NaCl.
ΔT = m Kf (solvent) (water) ΔT = (3.93 m) (1.86 C/m)ΔT = 7.3098 x 2 mol = 14.6Tf = 0 – 14.6 = -14.6
A 0.200 g sample of a nonvolatile yellow crystalline solid is dissolved in 18.2 g of benzene, producing a solution that freezes at 5.00oC. Find the molar mass of the yellow solid in g/mol.
Freezing Point of Benzene = 5.51 CKf Benzene = 4.90 C/mΔT = 5.51 – 5 = .51 C nonvolatile = 0.200 g benzene = 18.2 g ΔT = m Kf(solvent) (benzene) .51 C = m (4.90 C/m).104 mol /kg = m.200g (yellow solid) / 18.2g (Beneze) = .010989(.010989) x (1000/.104) = 106
A 3.90 g sample of a nonvolatile nonelectrolyte blue crystalline solid is dissolved in 35.3 g of acetic acid, producing a solution that bolis at 119.9oC. Find the molar mass of the blue solid in g/mol.
Boiling Point of Acetic Acid = 117.90 CKb Acetic Acid = 3.07 C/mΔT = 119.90 – 117.90= 2 C ΔT = m Kf(solvent) (benzene) 2 C = m (3.07 C/m).651 mol /kg = m3.90g (blue crystal) / 35.3g (Acetic Acid)= .11048(.11048) x (1000/ 0.651 (mol /kg) ) = 169.7 = 170
At a certain temperature the vapor pressure of pure benzene and pure toluene are 380 torr and 130 torr, respectively. If the mole fraction of benzene is 0.26, what is the total vapor pressure.
At a certain temperature the vapor pressure of pure benzene and pure toluene are 380 torr and 130 torr, respectively. If 7.58 mol benzene is mixed with 4.38 mol toluene, what is the total vapor pressure.