# Question 76a8d

May 11, 2015

You'd need to add 10 mL of hydrogen to that volume of air to reach a concentration of 300 ppm.

A hydrogen concentration of 1 ppm can be expressed as the ratio between $\text{1 L}$ of hydrogen and ${10}^{6} \text{L}$ of air

"1 ppm" = "1 L hydrogen"/(10^(6)"L air") = 10^(-6)

This means that you can get the concentration in ppm by multiplying the ration between the volume of hydrogen and the volume of air by ${10}^{6}$.

In your case, a concentration of 300 ppm will require a volume of hydrogen equal to

"ppm" = V_(H_2)/V_"air" * 10^(6) => V_(H_2) = ("ppm" * V_"air")/10^6

V_(H_2) = (300 * "36 L")/10^6 = "0.0108 L"#

Rounded to one sig fig, the number of sig figs given for 300 ppm, and expressed in mL, the answer will be

${V}_{{H}_{2}} = \textcolor{g r e e n}{\text{10 mL}}$