# Question 3379d

Mar 14, 2016

$3.3 \cdot {10}^{- 3} \text{g CO}$

#### 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, $\mu {\text{g m}}^{- 3}$, but that the answer must be expressed in grams per cubic feet, ${\text{g ft}}^{- 3}$.

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

$\text{1 g" = 10^6mu"g}$

and the other to take you meters to feet

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

This means that ${\text{1 ft}}^{3}$ will be equivalent to

$\text{1 ft"^3 = "1 ft" xx "1 ft" xx "1 ft}$

$\text{1 ft"^3 = "0.3048 m" xx "0.3048 m" xx "0.3048 m}$

${\text{1 ft"^3 = "0.028317 m}}^{3}$

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} V = l \times h \times w \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$l$ - the length of the prism
$w$ - its width
$h$ - its height

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.

$V = \text{9.0 ft" xx "14.5 ft" xx "18.8 ft}$

$V = {\text{2453.4 ft}}^{3}$

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

$2453.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{ft"^3))) * overbrace((1.3592 * 10^(-6)"g")/(1color(red)(cancel(color(black)("ft"^3)))))^(color(purple)("equivalent to 48"mu"g m"^(-3))) = color(green)(|bar(ul(color(white)(a/a)3.3 * 10^(-3)"g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.