Question #51727

It looks to me like you're trying to determine the mass of a given volume of a substance by using a #"m"^3 -> "cm"^3# conversion factor and the density of the substance.

On that note, I assume that your calculation looks like this

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = ?#

The first thing to note here is that the original measurement has #4# significant figures, the number of sig figs you have in its mantissa.

#7.379 * 10^(-4)" " -> " 4 sig figs: " {7, 3, 7, 9}#

Now, you're using a conversion factor to take you from cubic meters to cubic centimeters

#"1 m"^3 = 100^3 quad "cm"^3#

Because #"1 m"^3# is defined as being equivalent to #100^# #"cm"^3#, you can use as many significant figures as you want for this conversion factor.

In other words, #"1 m"^3# and #100^3# #"cm"^3# contain an infinite number of significant figures because they are defined that way.

Finally, you have the density of the substance, which is equal to #"19.3 g cm"^(-3)#, or #"19.3 g"# for every #"1 cm"^3#. This time, you have #3# significant figures in the mass of the substance present in #"1 cm"^3#.

#"19.3 g " -> " 3 sig figs: " {1, 9, 3}#

You can use the fact that density is defined as the mass of exactly one unit of volume to use an infinite number of sig figs for #"1 cm"^3#.

So you can say that in terms of the number of sig figs, your calculation comes down to

#"4 sig figs" xx "as many sig figs as you need"/"as many sig figs as you need" xx "3 sig figs"/"as many sig figs as you need"#

Since you're dealing with a multiplication, the answer must be rounded to the number of sig figs of your least precise measurement.

In this case, you have

#"3 sig figs " ("19.3 g") " " < " " "4 sig figs " (7.379 * 10^(-4))#

so the result must be rounded to #3# significant figures. This means that

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = "14,241.47 g"#

will be rounded to #3# sig figs to get--I'll express the answer in scientific notation

#7.379 xx 10^(-4) color(red)(cancel(color(black)("m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = color(darkgreen)(ul(color(black)(1.42 xx 10^(4) quad "g")))#