A) \[0.0716\]
B) \[1.30\times {{10}^{-5}}\]
C) \[1.25\times {{10}^{-2}}\]
D) 0.0555
Correct Answer: D
Solution :
[d] 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 \[{{H}_{2}}O=100mL\,{{H}_{2}}O\] \[=1000\,g\,{{H}_{2}}O(d=1\,g\,m{{L}^{-3}})\] \[\therefore \]Moles of \[{{H}_{2}}O=\frac{1000}{18}=55.5\]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) \[\therefore n=55.5\times 1.0\times {{10}^{-5}}\] \[=5.55\times {{10}^{-4}}mol=5.55\times {{10}^{-4}}\times 1000m\,\,\text{mol}\]\[=5.55\times {{10}^{-2}}\]millimole \[=0.0555\]millimole
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