Cliffs AP Chemistry, 3rd Edition

24.(B) The molarity of a solution multiplied by its volume equals the number of moles of

solute. In this case, 450 mL of 0.35 M Al(NO 3 ) 3 can be shown as

.(). .()

liter solution

mole Al NO liter solution mole Al NO

1

035

1

045

(^33) # = (^01633)
Al(NO 3 ) 3 is completely soluble, so there would be three times the number of moles of ni-
trate ions present in the solution because
Al NO( ) Al+()aq 3 NO ()aq
33
3
" + 3

Therefore, the number of moles of nitrate ions in the original solution would be 0.16 x 3 =

0.48.

The number of moles of nitrate ions needs to be brought up to 0.77 because the volume

did not change (it remained at 0.45 liter).

..
liter solution

moles NO liter solution

1

17

1

3 045

# = 3

• 
  
  
  
  • 
    
    
    
    77. mole of NO in final solution
    
    
    
    
  
  
  
  

The solution begins with 0.48 mole of nitrate ions and must end up with 0.77 moles of ni-

trate ions; therefore, the solution needs an additional 0.29 mole of nitrate ions:

^h077 048 029. -=.. mole NO needed 3 -

Calcium nitrate, Ca(NO 3 ) 2 , produces 2 moles of nitrate ions in solution for each mole of

solid calcium nitrate added to the solution. Therefore, because 0.29 mole of NO 3 – is

needed, you will need 0.29 / 2 = 0.15 mole of solid Ca(NO 3 ) 2.

25.(C) In examining the rate expression, note that B is first-order, so the rate is directly pro-

portional to the concentration of the reactant. Holding [A] constant, and doubling [B]

would double the rate.

26.(C) The term -d[B] /dtrepresents the rate of decrease in the concentration of B as time

elapses. For every mole of B that is lost on the reactant side,^1 ⁄ 2 x 4, or 2, moles of D are

gained on the product side over the same amount of time (dt).

27. (C) The rate constant is independent of the concentration of the reactants. However, kde-
    pends on two factors:
    
    
    • The nature of the reaction. “Fast” reactions typically have large rate constants.
    
    
    • The temperature. Usually kincreases with an increase in temperature.
      With all other variables held constant, choice (D) would reduce the rate of molecular colli-
      sions, but increasing the volume is analogous to decreasing the concentration.
      28.(D) The overall order of the reaction is the sum of the orders of the individual reactants.
      Here, [A]^2 [B]^1 = 2 + 1 = 3. For reactions with an overall order of 3, the rate is proportional
      to the cube of the concentration of the reactants. Reducing the volume by 1/3 effectively
      triples their concentration: 3(conc)3(order)= 27.
    
    
    
    

Part IV: AP Chemistry Practice Test