Question

review i
1 your ideas
you and a friend need to find
the density of a metal in the
lab. the metal has a weight of
23.2 grams and a volume of
8.593 ml. you calculate the
density as 2.67, and your friend
calculates the density as
2.69987199. which is the better
response? why?

Explanation:

Step1: Recall the rule of significant figures

In density calculation \(d=\frac{m}{V}\), the number of significant figures in the result is determined by the least number of significant figures in the given values. The mass \(m = 23.2\) grams has three significant figures, and the volume \(V=8.593\) mL has four significant figures.

Step2: Determine the number of significant figures for the density

According to the rule of significant - figures for division (\(d=\frac{m}{V}\)), the result should have the same number of significant figures as the value with the least number of significant figures in the division. Here, since \(m = 23.2\) (three significant figures) and \(V = 8.593\) (four significant figures), the density \(d=\frac{23.2}{8.593}\approx2.69\) (three significant figures).

Your calculation of \(2.67\) has three significant figures (assuming the last digit is due to proper rounding), while your friend's result \(2.69987199\) has nine significant figures which is more than the number of significant figures allowed by the input data.

Answer:

Your response of \(2.67\) (assuming proper rounding to three significant figures) is better. Because in a density calculation \(d = \frac{m}{V}\), the number of significant figures in the result is determined by the measurement with the least number of significant figures. The mass (\(23.2\) grams, three significant figures) limits the number of significant figures in the density result. Your friend's result of \(2.69987199\) has too many significant figures and does not follow the rules of significant - figure propagation in measurements.