Page 178 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 11] MOLARITY 167

EXAMPLE 11.15. (a) Calculate the concentration of Zn 2+ produced when excess solid zinc is treated with 100.0 mL of
6.000 M HCl. Assume no change in volume. (b) Repeat the problem with solid aluminum.
+
Ans. (a) Zn + 2H −→ Zn 2+ + H 2
6.000 mmol H + 1 mmol Zn 2+

100.0mL = 300.0 mmol Zn 2+
1mL 2 mmol H +
300.0 mmol Zn 2+
= 3.000 M Zn 2+
100.0mL
(b) 2Al + 6H −→ 2Al 3+ + 3H 2
+
+ 3+
6.000 mmol H 2 mmol Al
100.0mL = 200.0 mmol Al 3+
1mL 6 mmol H +
200.0 mmol Al 3+
= 2.000 M Al 3+
100.0mL

Solved Problems

INTRODUCTION
11.1. Which, if either, has more sugar in it, (a) a half cup of tea with one lump of sugar or (b) a whole cup of
tea with two lumps of sugar?
Ans. Two lumps are more than one; the whole cup has more sugar. Note the difference between quantity of sugar
and concentration of sugar.

11.2. Which, if either, of the following tastes sweeter, (a) a half cup of tea with one lump of sugar or (b)a
whole cup of tea with two lumps of sugar?
Ans. They both taste equally sweet, since their concentrations are equal.

MOLARITY CALCULATIONS
11.3. Calculate the molarity of each of the following solutions: (a) 4.00 mol solute in 2.50 L of solution,
(b) 0.200 mol solute in 0.240 L of solution, (c) 0.0500 mol solute in 25.0 mL of solution, and (d) 0.240
mol solute in 750.0 mL of solution.
4.00 mol 0.0500 mol 50.0 mmol
Ans. (a) = 1.60 M (c) = 2.00 M or = 2.00 M
2.50 L 0.0250 L 25.0mL
0.200 mol 0.240 mol 240 mmol
(b) = 0.833 M (d) = 0.320 M or = 0.320 M
0.240 L 0.750 L 750.0mL
(Note: moles per liter or millimoles per milliliter, not moles per milliliter)

11.4. Calculate the number of moles of solute in each of the following solutions: (a) 1.50 L of 0.800 M
solution, (b) 1.66 L of 0.150 M solution, (c) 45.0 mL of 0.600 M solution, and (d) 25.0 mL of 2.00 M
solution.

0.800 mol 0.600 mol
Ans. (a)1.50 L = 1.20 mol (c) 0.0450 L = 0.0270 mol
1L 1L
0.150 mol 2.00 mol

(b)1.66 L = 0.249 mol (d)0.0250 L = 0.0500 mol
1L 1L