Solutions Question 248

Question: Henry’s law constant for $ N_2 $ at 310 K is 82.35 kbar. $ N_2 $ exerts a partial pressure of 0.840 bar. If $ N_2 $ gas is bubbled through water at 293 K, then the number of millimoles of $ N_2 $ that will dissolve in 1 L of water is

Options:

A) $ 0.0716 $

B) $ 1.30\times {10^{-5}} $

C) $ 1.25\times {10^{-2}} $

D) 0.0555

Show Answer

Answer:

Correct Answer: D

Solution:

Henry’s law constant is in the unit of pressure, hence we use relation $ p=K_{H}{\chi_{N_2}} $

$ {\chi_{N_2}}=\frac{p}{K_{H}}=\frac{0.840,bar}{82.35\times 10^{3}bar}=1.0\times {10^{-5}} $ 1 L $ H_2O=100mL,H_2O $

$ =1000,g,H_2O(d=1,g,m{L^{-3}}) $

$ \therefore $ Moles of $ H_2O=\frac{1000}{18}=55.5 $ moles Let the number of moles of nitrogen be n Then $ {\chi_{N_2}}=\frac{n}{n+55.5}\approx \frac{n}{55.5} $ (since, n is very small)

$ \therefore n=55.5\times 1.0\times {10^{-5}} $

$ =5.55\times {10^{-4}}mol=5.55\times {10^{-4}}\times 1000mmol $

$ =5.55\times {10^{-2}} $ millimole $ =0.0555 $ millimole



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