Question

Henry’s law constant for the solubility of nitrogen gas in water at 298 K is $1.0\times {10}^{-5}atm$. The mole fraction of nitrogen in air is 0.8. The number of moles of nitrogen from air dissolved in 10 mol of water at 298 K and 5 atm pressure is

• A. $4.0\times {10}^{-4}atm$
• B. $4.0\times {10}^{-5}atm$
• C. $5.0\times {10}^{-4}atm$
• D. $4.0\times {10}^{-5}atm$

Answer 1

The correct option is A $4.0\times {10}^{-4}atm$

$P_{N_2}=5\times 0.8=4$
Applying Henry’s law, we get
$P_{N_2}=K_H\times X_{N_2}$
or $4={10}^5\times X_{N_2}$
$X_{N_2}=4\times {10}^{-5}$
$\frac{n_{nitrogen}}{n_{nitrogen+n_{H_{2}O}}}=4\times {10}^{-5}$
Or $\frac{n_{nitrogen}}{n_{nitrogen}+10}=4\times {10}^{-5}$
Or $n_{nitrogen}=4\times {10}^{-4}$