A) \[1.78\times {{10}^{-4}}\]
B) \[1.78\times {{10}^{-5}}\]
C) \[1.78\times {{10}^{-4}}\]
D) \[1.78\times {{10}^{-6}}\]
Correct Answer: B
Solution :
\[{{K}_{H}}=4.27\times {{10}^{5}}\]mm Hg (at 298 K) \[p=760\,\,mm\] Applying Henry's law, \[p={{K}_{H}}x\]where\[x=\]Mole fraction or solubility of methane. \[x=\frac{p}{{{K}_{H}}}=\frac{(760\,\,mm)}{(4.27\times {{10}^{5}}mm)}\] \[=1.78\times {{10}^{-5}}\]You need to login to perform this action.
You will be redirected in
3 sec