Question

4/20 an irregularly shaped piece of gold was lowered into a graduated cylinder holding a volume of water equal to 17 ml. the height of the water rose to 20 ml. if the mass of the gold was 27 g, what was its density? 10.5 g/ml 9 g/ml 6.5 g/ml 8 g/ml

Explanation:

Step1: Find the volume of gold

The volume of the gold is the change in water volume. Initial water volume is 17 mL, final is 20 mL. So volume \( V = 20 - 17 = 3 \) mL.

Step2: Calculate density

Density formula is \(
ho=\frac{m}{V} \), where \( m = 27 \) g and \( V = 3 \) mL. So \(
ho=\frac{27}{3}=9 \) g/mL.

Answer:

B. 9 g/mL