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Chapter 1

Q. 1.7

Consider two samples of palladium (Pd), an element used in automobile catalytic converters. Sample A is a cylindrical bar with a mass of 97.36 g. The bar is 10.7 cm high and has a radius of 4.91 mm. Sample B is an irregular solid with a mass of 49.20 g. A graduated cylinder has 10.00 mL of water. When sample B is added to the graduated cylinder, the volume of the water and the solid is 14.09 mL. Calculate the density of each sample.

Sample A:

ANALYSIS
mass (97.36 g), radius, r (4.91 mm), height, h (10.7 cm) Information given:
density of Pd Asked for:

STRATEGY

1. Recall the formula to obtain the volume of a cylinder.

V = πr²h

2. Substitute into the definition of density.

d = \frac{mass}{V}

Sample B:

ANALYSIS
mass: (49.20 g)
volume of water before Pd addition: (10.00 mL)
volume of water and Pd: (14.09 mL)
Information given:
density of Pd Asked for:

STRATEGY

1. The volume of the Pd is the difference between the volume of the Pd and water and the volume of the water alone.
2. Substitute into the definition of density.

d = \frac{mass}{V}

Step-by-Step

Verified Solution

Sample A:

  V = πr²h = π\left(4.91  mm  ×  \frac{1  cm}{10 mm}\right)² × 10.7 cm = 8.10 cm³ V
d = \frac{mass}{V} = \frac{97.36  g}{8.10  cm³} = 12.0 g/cm³ d

Sample B:

V =  V_{H_{2}O  +  Pd}  – V_{H_{2}O} + 14.09 mL – 10.00 mL = 4.09 mL V
d = \frac{mass}{V} = \frac{49.02  g}{4.09  mL } = 12.0 g/mL d

END POINT

The units for density are g/cm³ and g/mL. Since 1 cm³ = 1 mL, these can be used interchangeably.