Question

A pure gold metal bar displaces 0.82 L of water. What is its mass in kilograms? (The density of gold is 19.3 g/cm³.)

a) 0.82 kg
b) 15.77 kg
c) 19.3 kg
d) 15.77 g

Answer 1

The mass of a pure gold bar that displaces 0.82 liters of water is approximately 15.77 kg, given the density of gold is 19.3 g/cm³.

Step-by-step explanation:

To determine the mass of a pure gold metal bar that displaces 0.82 L of water, we use the density of gold which is given as 19.3 g/cm³. First, we will convert the volume from liters to cubic centimeters, since there are 1000 cubic centimeters in a liter, the volume of the gold bar in cubic centimeters is 0.82 L × 1000 cm³/L = 820 cm³. Then we can calculate the mass of the gold in grams using the density formula: mass = density × volume. This gives us:

mass = 19.3 g/cm³ × 820 cm³

The mass of the gold bar is therefore 15826 g, or 15.826 kg when converted to kilograms (since 1 kg = 1000 g). Thus, the correct answer is (b) 15.77 kg, which is approximately the calculated mass of 15.826 kg after rounding to two decimal places.