# Question 51727

Jan 27, 2018

The answer must be rounded to three significant figures.

#### Explanation:

It looks to me like you're trying to determine the mass of a given volume of a substance by using a ${\text{m"^3 -> "cm}}^{3}$ 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.

$7.379 \cdot {10}^{- 4} \text{ " -> " 4 sig figs: } \left\{7 , 3 , 7 , 9\right\}$

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

${\text{1 m"^3 = 100^3 quad "cm}}^{3}$

Because ${\text{1 m}}^{3}$ is defined as being equivalent to 100^ ${\text{cm}}^{3}$, you can use as many significant figures as you want for this conversion factor.

In other words, ${\text{1 m}}^{3}$ and ${100}^{3}$ ${\text{cm}}^{3}$ 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 ${\text{19.3 g cm}}^{- 3}$, or $\text{19.3 g}$ for every ${\text{1 cm}}^{3}$. This time, you have $3$ significant figures in the mass of the substance present in ${\text{1 cm}}^{3}$.

$\text{19.3 g " -> " 3 sig figs: } \left\{1 , 9 , 3\right\}$

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 ${\text{1 cm}}^{3}$.

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

$\text{4 sig figs" xx "as many sig figs as you need"/"as many sig figs as you need" xx "3 sig figs"/"as many sig figs as you need}$

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

$\text{3 sig figs " ("19.3 g") " " < " " "4 sig figs } \left(7.379 \cdot {10}^{- 4}\right)$

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

$7.379 \times {10}^{- 4} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{m"^3))) xx (100^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("m"^3)))) xx ("19.3 g")/(1 color(red)(cancel(color(black)("cm"^3)))) = color(darkgreen)(ul(color(black)(1.42 xx 10^(4) quad "g}}}}$