Solutions to
Additional Problems with Chapter 4
1a. 250 mL of 0.100 M K4Fe(CN)6
250 mL X (1 L/1000 mL) X (0.100 mol/1 L) X (368.365 g/1 mol) = 9.21 g
1b. 250 mL of 2 ppm Na+ from NaCl
since this is a dilute, aqueous solution, it can be assumed that ppm = mg/L
250 mL X (1 L/1000 mL) X (2 mg Na+/1 L) X (1 mmol Na+/22.99 mg) X (1 mmol NaCl/1 mmol Na+) X (58.44 mg/1 mmol NaCl) X (1 g/1000 mg) = 1.27 X 10-3 g NaCl
1c. 250 mL of 2.50 ppb Na+ from Na2SO4
since this is a dilute, aqueous solution, it can be assume that ppb = μg/L
250 mL X (1 L/1000 mL) X (2.05 μg Na+/1 L) X (1 g/106 μg) X (1 mol Na2SO4 /2 mol Na+) X (142.05 g/1 mol) = 1.58 X 10-6 g Na2SO4
2. Dilution so use MV = MV
(0.620 M)V = (1000 mL)(0.500 M)
V = 806 mL
So measure 806 mL of 0.620 M acid into a 1 L volumetric flask and dilute to the mark. (Since the acid is not diluted by much here, adding acid to water should not be a problem.)
3a. 7.07 + 6.5 = 13.57 = 13.6
e = [(0.03)2 + (0.4)2] ½ = 0.4011
answer = 13.6±0.4
3b. 456.57/915.472 = 0.49873
%e = [(0.06/456.57)2 + (0.004/915.472)2]1/2 = 1.31 X 10-4
e = (1.31 X 10-4)(0.49873) = 6.557566 X 10-5
answer = 0.49873±0.00007
3c. the math in the parenthesis
(81.32)(0.1399)/(-3.12) = -3.64636 = -3.65
%e = [(0.09/81.32)2 + (0.0002/0.1399)2 + (0.01/3.12)2]1/2 = 3.67987 X 10-3
e = (3.67987 X 10-3)(3.65) = 0.0134
outside the parenthesis
-3.65 – 22.3323 = -25.98
e = [(0.0134)2 + (0.0001)2]1/2 = 0.0134
answer =
-25.98±0.01
4. average = 2.84
sample standard deviation = 0.12