A gold-colored object found in a cave has a mass of 8.48 g and a volume of 5 x 1

Question

A gold-colored object found in a cave has a mass of 8.48 g and a volume of

5 x 10-1 mL. Is this information sufficient to positively identify the sample as pure gold, whose density is 19.3 g/mL?

a. yes, because the sample has a density of 16.96 g/mL, which is close enough

b no, because the density to two significant figures is 17 g/mL

c. no, the volume needs to be measured much more precisely to make the comparison

d. b and c

I know the answer is C., and I know the density comes out to 16.96 g/mL, but what I do not understand is why .5 would not be accurate enough. Thanks.

Explanation / Answer

Ans. When representing a value as decimal, the last digit is always uncertain.

For example, you go to a grocery shop and put a coin on the eight scale thereby. If you ever noticed, the last digit fluctuates a bit before coming to a stable reading. So, the last digit if always uncertain. If not yet tried, you should give a try seeing the variation in mass while weighing.   

Consider an example-

I. 0.5 mL means that the last digit “5” is uncertain, that is it could be “4” or “6”. So, the actual volume may lie in the range 0.4 – 0.6 mL.

So, possible values of density would range between-

            a. 8.48 g/ (0.4) mL = 21.2 g/ mL                , and

            b. 8.48 g/ (0.6) mL = 14.13 g/ mL              , and

So, with the value of 0.5, the density of given material could lie anywhere in between 14.13 g/ mL - 21.2 g/ mL.

Therefore, it’s very important to get a more accurate value of volume of the material.

II. 0.53 mL means that the last digit “3” is uncertain. That is, the actual volume lie in the range 0.52 -0.54 mL.

Compare I and II, greater is the number of digits after decimal, greater is the accuracy and lesser is the possible variation or error in measurements.

Therefore, when it comes to check the purity of gold, both the mass and volume measurements must be very accurate. For example, a volume 0.500 mL shall give more accurate value.

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