A) 0.0555
B) \[55.55\times {{10}^{-5}}\]
C) \[55.55\times {{10}^{-3}}\]
D) \[5.55\times {{10}^{-5}}\]
Correct Answer: B
Solution :
\[P={{K}_{H}}\times X\] \[x=\frac{p}{{{K}_{H}}}=\frac{1}{100\times {{10}^{3}}}=1\times {{10}^{-5}}\] Mole fraction = \[\frac{Moles\,of\,gas}{Total\,moles}\] Moles of \[{{H}_{2}}O=~\frac{1000}{18}=\text{ }55.55\] \[(\because 1L=1000g)\] Mole fraction= \[\frac{x}{x+55.55}\]\[\left( 55.55>>>x \right)\] \[\therefore \] \[{{10}^{-5}}=\frac{x}{55.55}\] or \[x\,\,=\,\,55.55\times {{10}^{-\,5}}\]
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