# 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?

##### 1 Answer

#### 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 **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

The **volume** of a rectangular prism is given by the formula

#color(blue)(V = l xx w xx 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*.

#"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 = "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_(CO) = color(green)(3500color(white)(a)mu"g")#