# Question 73137

May 10, 2015

The concentration of triclosan will be $2.5 \cdot {10}^{- 2} \text{ppb}$.

You can write a concentration expressed in parts per billion, or ppb, as 1 gram of solute, in your case triclosan, per 1 billion grams of solvent, in your case water.

A 1 ppb concentration means that you have 1 gram of triclofan in 1 billion grams of water

"1 ppb" = "1 g triclosan"/(10^9"g water") = 10^(-9)

This means that if you multiply the ratio between the mass of solute and the mass of solution by ${10}^{9}$, you'll get the concentration in ppb.

You can convert volume to grams by using water's density

1.5cancel("L") * "1000 g"/(1cancel("L")) = 1.5 * 10^(3)"g water"

Expressed in grams, the mass of triclosan will be

38cancel("ng") * ("1 g")/(10^(9)cancel("ng")) = 38 * 10^(-9)"g"#

The ratio between the mass of triclosan and the mass of the solution will be

$\left(38 \cdot {10}^{- 9} \cancel{\text{g"))/(1.5 * 10^(3) * cancel("g}}\right) = 2.533 \cdot {10}^{- 11}$

Multiply this by ${10}^{9}$ to get concentration in ppb

$2.533 \cdot {10}^{- 11} \cdot {10}^{9} = \textcolor{g r e e n}{2.5 \cdot {10}^{- 2}}$

(the answer is rounded to two sig figs).