NAME:__________________________
Chemistry 150
Quiz 6
November 10, 2003
1. If 3.18 g of BaCl2 is dissolved in enough solvent to make 500.0 mL of solution, what
is the molarity? (2 points)
FW(BaCl2) = 208.23 g/mol
3.18 g BaCl2 • 1 mol BaCl2/208.23 g BaCl2 = 0.01527 mol BaCl2
0.01527 mol BaCl2/0.500 L solution = 0.031 mol/L = 0.031 M BaCl2
2. Calculate the w/v percentage of each of these solutes: (2 points each):
a. 623 mg of casein in 15.0 mL of milk
%(w/v) =
623 mg casein
15.0 mL sol
1 g casein
x 100% = 4.15% w/v casein
1000 mg casein
b. 3.25 g of sucrose in 186 mL of coffee
3.25 g sucrose
c. %(w/v) =
x 100% = 1.75% w/v sucrose
186 mL sol
3. The following reaction is exothermic:
2NO(g) + Br2(g) ↔ 2NOBr(g)
After it reaches equilibrium, we add a few drops of Br2.
a. What will happen to the equilibrium? (1 point)
The equilibrium shifts to the right, or products, because reactant was added.
b. What will happen to the equilibrium constant? (1 point)
The equilibrium constant is not affected, it stays unchanged.
4. Which of the following molecules cannot engage in hydrogen bonding: (1 point,
circle one answer only.)
a.
b.
c.
d.
e.
H2O
CH4
NH3
HCl
HI
5. For the reaction below, what is the expression for K? (1 point, circle one answer only.)
2H2(g) + 2FeO(s) ↔ 2Fe(s) + 2H2O(g)
a.
b.
c.
d.
e.
K = [Fe]2[H2O]2/[H2]2[FeO]2
K = [H2O]2/[H2]2[FeO]2
K = [Fe]2/[H2]2[FeO]2
K = [H2O]2/[H2]2
K = [2Fe]2[2H2O]2/[2H2]2[2FeO]2
POINTS:________( of 10)