Calculating the mass of acceptable carbon monoxide in a room The acceptable concentration of $\ce{CO}$ in the air is $10\:\mathrm{mg/m}^3$. In
  a room that is $19\:\mathrm{m\times4.0\:m\times25\:m}$, what is the acceptable mass in
  kilograms of $\ce{CO}$?


The room has a volume of $1900\:\mathrm{m}^3$. Since there are $10\:\mathrm{mg}$ of $\ce{CO}$ per $\mathrm{m}^3$, there are $19000\:\mathrm{mg}$ of $\ce{CO}$ in this room. That’s $19\:\mathrm{kg}$.

But the book says the answer is $1.9 \times 10^{-2}\:\mathrm{kg}$. How come?
You have to be careful with your conversions - you are missing a step in your conversions.

You have 19000 mg

As 1 mg = 10-3 g  and
1 g = 10-3 kg

So:

19000 mg = 19 g = 1.9 x 10-2 kg