1. A solution is prepared by mixing 1.0 grams of benzene (C₆H₆) in 100 g of water to create a solution total volume of 100 ml. Calculate the molarity, mass percent, mole fraction, and molality of benzene in the solution.
2. Which ion would you expect to be more heavily hydrated?
   1. K⁺, Ca²⁺
   2. Na⁺, K⁺
   3. Cu²⁺, Cu⁺
3. Assuming the CO₂ partial pressure in air above a lake at sea level is 4.0 x 10^-4 atm, what is the equilibrium concentration of CO₂ in the lake at 25^oC? (Henry’s law constant is 32 L ^. atm / mole)
4. A cocktail was prepared by mixing 20.0g ethanol to 150 g water at 25^oC. Pure water has a vapor pressure of 23.76 torr. What would be the new vapor pressure?
5. 10 grams of salt (NaCl) is added to 100 mL of water. What are the new freezing and boiling points? (K_b = .51 ^oC kg/mol, K_f = .1.86 ^oC kg/mol)
6. Calculate the pressure needed to prevent osmosis when 10.0 g CaCl₂ is added to 100 mL of water?

1.

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(0.128 moles0 / (0.100 L) = 1.28 Moles/Liter

total mass = 1.0 g benzene + 100 g water = 101 g

1.0 grams benzene / (total mass)

1.0 g benezine / 101 g * 100 = .99 percent benzene

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(100 g water) / (16 g/mol) = 6.25 moles water

(.0128 moles benzene) / (.0128 moles benzene + 6.25 moles water) = .20 % benzene

molality

(1.0 grams benzene) / (78 g/mol) = .0128 moles benzene

(.0128 moles benzene) / (.1 kg water) = 1.28 mole/Kg

2.

1. Ca²⁺ should be more heavily hydrated because there is a greater attractive force due to the extra charge making D H₃ more negative. D H₁ and D H₂ are going to be similar because these are similarly sized ions.
2. Na⁺ should be more heavily hydrated because it is the smaller ion, which will decrease D H₁ and D H_2.
3. Cu²⁺ should be more heavily hydrated because of the same reasons as in part a.

3.

P_CO2 = k_CO2C_CO2

C_CO2 = P_CO2 / k_CO2

C_CO2 = 4.0 E^-4 atm / 32 L atm/mol

C_CO2 = 1.2 E^-5 mol/L

4.

P = c _solvent P^o

20 g ethanol / 46 g/mol = .44 mol ethanol

150 g water / 16 g.mol = 9.4 mol water

c _solvent = 9.4 mol water / (9.4 mol water + .44 mol ethanol)

c _solvent = .96

P = 0.96 * 23.76 mm Hg = 23 mm Hg

5.

m_solute = moles NaCl / Kg water

m_solute = (10 g / 58.5 g/mol) / .10 kg water

m_solute = 1.7 mol/kg

T_b = T^o_b + K_b m_solute

T_b = 100 + .51* 1.7 * 2

T_b = 102 ^oC

T_f = T^o_f - K_b m_solute

T_f = 0 – 1.86* 1.7 * 2

T_f = -6.3 ^oC

Note that i in this case is 2

6.

1.0 g CaCl₂/ 111.0 g/mol = .00900 mol CaCl₂

.00900 mol CaCl₂ / .100 L = .0900 M CaCl₂

p = iMRT

p = 3*.0900*.08206 * 298

p = 6.60 atm