Solutions Question 75
Henry’s law constant for $ N_2 $ at 293 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
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Answer:
Correct Answer: D
Solution:
Henry’s law constant is in the unit of pressure, hence we use relation $ p=K_{H} \cdot \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=100,mL,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
BETA
