An irregular 77.7 g piece of silver is placed into a graduated cylinder .....?

Question

Hi, heres my question:

An irregular 77.7 g piece of silver is placed into a graduated cylinder that has 15.0 mL of water in it. The new water level is 22.4 mL. What is the density of the silver?

To get the correct answer for this question, you must pay attention to "How many significant figures result after the subtraction?"

0.095 g/cm3

3.5 g/cm3

10.5 g/cm3

1.0x101 g/cm3

None of these answers

Answer 1

22.4mL - 15.0 mL = 7.4 mL  (1cm3/1mL) = 7.4cm3 is the volume of the sample.

77.7g / 7.4cm3 = 10.5g/cm3

Answer 2

The volume of the silver piece is 7.4 ml (or cubic centimetre) since this is how much the level of water changed.

With a mass of 77.7 g, the density is thus 77.7/7.4 = 10.5 g/cm^3

But, since we only have 2 significant digits for the volume, the 0.5 bit is uncertain, so the actual density cannot be known to better than between 10.4 and 10.6 (taking the volume to be 7.3 or 7.5, to account for the lack of precision in the volume measurmement) , i.e. 10.5 +/- 0.1.

That could point to the right answer being 1.0 * 10^1, which is just 10...