Answer
If $P_{N_{2}}= 1520\,mmHg$, then
$S_{N_{2}}=1.3\times10^{-3} mol/L $
When $P_{N_{2}}=20\,mmHg$,
$S_{N_{2}}=1.7\times10^{-5} mol/L $
Work Step by Step
Henry's law constant for the given case is
$k_{H}= 8.4\times10^{-7}mol\,L^{-1}\, mmHg^{-1}$
According to Henry's law,
$S_{g}= k_{H}P_{g}$ where $S_{g}$ is the solubility of the gas and $P_{g}$ is the partial pressure of the gas.
If $P_{N_{2}}= 1520\,mmHg$, then
$S_{N_{2}}=8.4\times10^{-7}mol\,L^{-1}\, mmHg^{-1}\times 1520\, mmHg=1.3\times10^{-3} mol/L $
When $P_{N_{2}}=20\,mmHg$,
$S_{N_{2}}=8.4\times10^{-7}mol\,L^{-1}\, mmHg^{-1}\times 20\, mmHg=1.7\times10^{-5} mol/L $
