# The concentration of carbon monoxide in an urban apartment is 48 µg/m^3. What mass of carbon monoxide in grams is present in a room measuring 11.0 ft by 11.5 ft by 20.5 ft?

Jan 29, 2016

$3500 \mu \text{g}$

#### Explanation:

The idea here is that you need to use the information provided by the problem to find the volume of the room.

Since you are told that the concentration of carbon monoxide in a typical apartment is equal to $48 \mu {\text{g/m}}^{3}$, finding the volume of the room in cubic meters will allow you to determine how much carbon monoxide it contains.

Now, you can treat the room as a rectangular prism of length $l$, width $w$, and height $h$.

The volume of a rectangular prism is given by the formula

$\textcolor{b l u e}{V = l \times w \times h}$

You can use the conversion factor that exists between feet and meters to find the dimensions of the room in meters, then calculate the volume in cubic meters.

$\text{1 ft " = " 0.3048 m}$

The volume of the room will thus be

V = (20.5color(red)(cancel(color(black)("ft"))) * "0.3048 m"/(1color(red)(cancel(color(black)("ft"))))) xx (11.5color(red)(cancel(color(black)("ft"))) * "0.3048 m"/(1color(red)(cancel(color(black)("ft"))))) xx (11.0color(red)(cancel(color(black)("ft"))) * "0.3048 m"/(1color(red)(cancel(color(black)("ft")))))

$V = {\text{73.43 m}}^{3}$

This means that the room will contain

73.43color(red)(cancel(color(black)("m"^3))) * overbrace( (48color(white)(a)mu"g")/(1color(red)(cancel(color(black)("m"^3)))))^(color(purple)("given concentration")) = 3524.64mu"g"

Rounded to two sig figs, the number of sig figs you have for the concentration of carbon monoxide, the answer will be

${M}_{C O} = \textcolor{g r e e n}{3500 \textcolor{w h i t e}{a} \mu \text{g}}$