# 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?