5 Steps to a 5 AP Chemistry

192  Step 4. Review the Knowledge You Need to Score High

3. C—The net ionic equation is:

The strontium nitrate solution contains:

(70.0 mL)(0.20 mol Sr(NO 3 ) 2 /1000 mL)

(1 mol Sr^2 +/1 mol Sr(NO 3 ) 2 )

=0.014 mol Sr^2 +

The sodium sulfate solution contains:

(30.0 mL) (0.10 mol Na 2 SO 4 /1000 mL) (1 mol

SO 42 −/1 mol Na 2 SO 4 ) = 0.0030 mol SO 42 −

The strontium and sulfate ions react in a 1:1

ratio, so 0.0030 mol of sulfate ion will combine

with 0.0030 mol of strontium ion leaving 0.011 mol

of strontium in a total volume of 100.0 mL. The

final strontium ion concentration is:

4. B—A (nitrous acid) and D (acetic acid) are weak
   acids, and E (ammonia) is a weak base. Weak
   acids and bases are weak electrolytes. C (ethanol)
   is a nonelectrolyte. Potassium nitrate (B) is a
   water-soluble ionic compound.


5. B—The number of moles of chloride ion needed is:

(400 mL)(1.0 mol Cl−/1000 mL) =0.40 mol Cl−

The initial number of moles of chloride ion in

the solution is:

(400 mL)(0.30 mol MgCl 2 /1000 mL)(2 mol

Cl−/mol MgCl 2 ) =0.24 mol Cl−

The number of moles needed =[(0.40 – 0.24)

mol Cl−] (1 mol CaCl 2 /2 mol Cl−) =0.080 mol

6. D—Both of the acids are strong acids and yield

1 mol of H+each. Calculate the number of moles

of H+produced by each of the acids. Divide the

total number of moles by the final volume.

(30.0 mL)(0.50 mol H+/1000 mL)

+(70.0 mL)(1.00 mol H+/1000 mL)

=0.085 mol H+

7. A—To calculate the molality of a solution, both

the moles of solute and the kilograms of solvent

are needed. A liter of solution would contain a

known number of moles of solute. To convert

this liter to mass, a mass to volume relationship

(density) is needed.

8. E—The mole fraction may be determined by
   dividing the vapor pressure of the desired sub-
   stance by the sum of all the vapor pressures.

87 mm Hg/(87 mm Hg +170 mm Hg) =0.34

9. C—To produce a molar solution of any type, the
   final volume must be the desired volume. This
   eliminates answers D and E. B involves mass of
   water instead of volume. A calculation of the
   required mass will allow a decision between A
   and B.

(3.0 L)(0.20 mol K 3 PO 4 /L)(212 g K 3 PO 4 /1 mol

K 3 PO 4 ) =130 g K 3 PO 4

10. C—A 5.2 molal solution has 5.2 mol of methyl
    alcohol in one Kg (1000 g) of water. The moles
    of water are needed.

(1000 g H 2 O)(1 mol H 2 O/18.0 g H 2 O)

=55.6 mol H 2 O

The mole fraction may now be determined.

(5.2 mol CH 3 OH)/(5.2 +55.6) mol =0.086

11. B—The boiling point depends on the boiling-
    point elevation (a colligative property). All col-
    ligative properties depend on the concentration
    of particles present. C is a nonelectrolyte, thus,
    the concentration of the particles is 0.20 M. All
    of the other substances are strong electrolytes.
    The concentration of particles in each of these
    may be determined by multiplying the molarity
    given by the van’t Hoff factor. Each of these is cal-
    culated as follows:

A: 2 ×0.10 =0.20 B: 4 ×0.10 =0.40

D: 2 ×0.10 =0.20 E: 2 ×0.10 =0.20

12. D—

(5.0 mole MgSO 4 /1000 mL)(100.0 mL)

(120.4 g MgSO 4 /mol MgSO 4 )

=60. g MgSO 4

(0.085 mol H )(1000 mL)

(100.0 mL)(1 L)

+

=08.55MH+

(0.011 mol Sr )(1000 mL)

(100.0 mL)(1 L)

2 +

=011.MMSr^2 +

Sr^2 +−(( ()aq) + SO^24 aq)→SrSO s 4