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Answer 1 · 2015-05-11T21:08:52

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 $"1 L"$ of hydrogen and $10^(6)"L"$ 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

$V_(H_2) = color(green)("10 mL")$