1. ?H_solution = ?H_solute + ?H_solvent + ?H

   mixing_{230 J/mol = ?H}solute _{+ 108 J/mol + - 47 J/mol}

   ?H_solute = 230 J/mol - 108 J/mol +47 J/mol

   ?H_solute = 169 J/mol

2. Solution is only ideal when ?H_solution = 0

   -219 J/mol does not equal 0

   Therefore, the solution is non-ideal.

3. (c)
4. (a)     MgCl₂ in CCl₄  No (Ionic compound (ions in solution) and  non-polar)    (b)  NaCl in water  Yes (polar and polar)  

(c)     Pentane, C₅H₁₂, in CCl₄  Yes (Polar and polar)      (d)  methanol, CH₃OH,  in water Yes (Polar and polar)

(e)  Vegetable oil in mineral oil Yes (non-polar and non-polar)

5. (a)

6. a) hexane, toluene, or CCl₄ (dispersion forces); b) same as a; c) water, acetone, methanol, ethanol (dispersion forces, ion-dipole); d) same as a and b

7. When the temperature of the solution (water as solvent and oxygen gas as the solute) is raised the solubility of the oxygen in water is lowered so the gas bubbles out of solution. The result is much less oxygen in the water solution. When the fish is put in this solution with to little oxygen, it can not survive.

8. C = kP = 3.7 x 10^-4 M/atm · 1.0 atm = 3.7x10^-4 M 

9. First convert the mass of water to kg.  

Next you will convert the grams of glucose to moles.

Calculate the molality.

10. Solution

P pure H₂O = 23.8 mm Hg

To solve for the mole fraction, you must first convert the 2 L of water into moles:

1 L = 1000 mL = 1000 g

Knowing this, you can convert the mass of water (2000 g) into moles:

2000 g / 18.02 g (molar mass of water) = 110.9 moles H₂O

Solve for the mole fraction, x_H2O:

x_H2O = moles H₂O / total moles
= 110.9 moles / 110.9 + 1.5 moles = 0.979

Finally, apply Raoult's Law

P_Kool-Aid = x_H2O P_{pure H2O} = (.979)(23.8 mm Hg) = 23.3 mm Hg

11. 

Solution

If x_benzene = 0.6, than x_toluene = 0.4 because 1 - 0.6 = 0.4.

Now that we know the mole fractions and vapor pressures, this problem is a cinch.

P_benzene = x_benzeneP_benzene = (0.6)(95.1 mm Hg) = 57.1 mm Hg
P_toluene = x_tolueneP_toluene = (0.4)(28.4 mm Hg) = 11.4 mm Hg

The total vapor pressure is simply the sum of the partial pressures:

P_total = P_benzene + P_tuolene = 57.1 mm Hg + 11.4 mm Hg = 68.5 mm Hg

12. The K_bH2O= 0.512°C/m

so

?T_b = (0.0512°C/m)(1.25m) = 0.640°C

The solution would boil at a temperature that is 0.640°C higher than pure water (boiling point elevation). The normal boiling point of pure water is 100°C so we would predict this solution to boil at 100°C+0.640°C = 100.640°C.

13.

25°C + 273.15 = 298.15K

500.0 mL x (1L/1000mL) = 0.5000 L

8.92 g KBr (1 mol KBr/199.01 g KBr) = 0.0750 mol KBr

Calculate the Molarity of the solution:

M = 0.0750 mol KBr/ 0.5000 L = 0.150 M KBr

? = MRT = (0.150 M) (0.08206 L*atm/mol*K) (298K) = 3.67

i = actual osmotic pressure/osmotic pressure that would be observed if no ionic dissociation occured = 6.97/3.67 = 1.899

We expect i to be 2 if the KBr completely dissociates so this value seems reasonable.

14. (c)