Page 48 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 2] MATHEMATICAL METHODS IN CHEMISTRY 37

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2.90. Calculate the mass of a gold sphere (density = 19.3g/cm ) (a) of radius 3.00 cm and (b) of diameter 3.00 cm.
4
V = πr 3
3
4 d
3
Ans. (a) V = π(3.00 cm) = 113 cm 3 (b) r = = 1.50 cm
3 2
19.3g 4

m = 113 cm 3 = 2180 g = 2.18 kg V = π(1.50 cm) = 14.1cm 3
3
1cm 3 3

19.3g
m = 14.1cm 3 = 272 g
1cm 3
2.91. How far can Mary drive her car on $50.00 if the car goes 25.4 mi/gal and gasoline costs $1.899 per gallon?

1gal 25.4mi
Ans. 50.00 dollars = 669 mi
1.899 dollars 1gal
2.92. (a) What is the mass of 18.6 mL of mercury? (Density = 13.6 g/mL) (b) What is the volume of 18.6 g of mercury?

13.6g
Ans. (a)18.6mL = 253 g
1.00 mL
18.6g
(b) = 1.37 mL
13.6 g/mL
(a) When we multiplied the mass per milliliter by the volume, the units mL canceled, and we were left with
grams, which is the unit of mass we wanted. (b) When we divided the mass by the mass per milliliter, the
units grams canceled, and we were left with mL, which is the unit of volume we wanted. If we had multiplied
in part (b) or divided in part (a), the units would not have canceled, and we would have obtained a result that
was not appropriate to the problem. The units helped us decide whether to multiply or divide.
Instead of dividing by a ratio we can do the same thing by inverting the ratio and multiplying by it.
2.93. Calculate the density of 150.0-g rectangular block with length 15.00 cm, width 2.50 cm, and thickness 1.50 cm.

Ans. 150.0g = 2.67 g/cm 3
(15.00 cm)(2.50 cm)(1.50 cm)

2.94. Using the data of Problem 2.67, determine whether a cube of metal 1.50 cm on each side is mercury, gold, or lead.
The mass of the cube is 65.2 g.
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Ans. The volume of a cube is equal to its length, cubed, or (1.50 cm) = 3.375 cm .
m (65.2g) 3
d = = = 19.3 g/cm
V (3.375 cm )
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The metal is gold.
2.95. Commercial sulfuric acid solution is 96.0% H 2 SO 4 in water and has a density of 1.86 g/mL. How many grams of
H 2 SO 4 are there in 37.1 mL of the commercial solution?

1.86 g solution 96.0gH 2 SO 4
Ans. 37.1 mL solution = 66.2gH 2 SO 4
1 mL solution 100 g solution
↑ ↑ ↑
Quantity given Density of solution Percent
2.96. What mass of H 2 SO 4 is present in 41.7 g of solution which is 96.0% H 2 SO 4 in water?

96.0gH 2 SO 4
Ans. 41.7 g solution = 40.0gH 2 SO 4
100 g solution