# Question #3379d

##### 1 Answer

#### Answer:

#### Explanation:

This is a great example of a **unit conversions** problem.

Notice that the concentration of carbon monoxide is given in *micrograms per cubic meter*, *grams per cubic feet*,

This means that you're going to have to use *two* conversion factors, one to take you from *micrograms* to *grams*

#"1 g" = 10^6mu"g"#

and the other to take you *meters* to *feet*

#"1 ft " = " 0.3048 m"#

This means that

#"1 ft"^3 = "1 ft" xx "1 ft" xx "1 ft"#

#"1 ft"^3 = "0.3048 m" xx "0.3048 m" xx "0.3048 m"#

#"1 ft"^3 = "0.028317 m"^3#

The dimensions of the room will then help you find its *volume*. As you know, the volume of a **rectangular prism** is given by

#color(blue)(|bar(ul(color(white)(a/a)V = l xx h xx wcolor(white)(a/a)|)))" "# , where

One approach to have here is to convert the concentration of carbon monoxide from micrograms per cubic meter to grams per cubic feet by using the aforementioned conversion factors

#48 color(red)(cancel(color(black)(mu"g")))/(color(red)(cancel(color(black)("m"^3)))) * (0.028317color(red)(cancel(color(black)("m"^3))))/("1 ft"^3) * "1 g"/(10^6color(red)(cancel(color(black)(mu"g")))) = 1.3592 * 10^(-6)"g ft"^(-3)#

Next, use the dimensions of the room to find its volume.

#V = "9.0 ft" xx "14.5 ft" xx "18.8 ft"#

#V = "2453.4 ft"^3#

This means that the mass of carbon monoxide in this particular room will be

#2453.4 color(red)(cancel(color(black)("ft"^3))) * overbrace((1.3592 * 10^(-6)"g")/(1color(red)(cancel(color(black)("ft"^3)))))^(color(purple)("equivalent to 48"mu"g m"^(-3))) = color(green)(|bar(ul(color(white)(a/a)3.3 * 10^(-3)"g"color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.