A) 0.555
B) 5.55
C) 0.0555
D) 55.5
Correct Answer: A
Solution :
| \[{{\operatorname{K}}_{H}}=100kbar=1{{0}^{5}}bar,\] |
| \[\operatorname{P}={{K}_{H}}\times {{x}_{A}}\] |
| \[{{\operatorname{x}}_{A}}=\frac{P}{{{K}_{H}}}=\frac{1}{100\times 1{{0}^{3}}}=1{{0}^{-5}}\] |
| Weight of water \[=1000\text{ }g\] |
| Moles of water \[=\frac{1000}{18}=55.5\] |
| \[\left( \because 1\operatorname{mL}=1000g \right)\] |
| Mole fraction \[=1{{0}^{-5}}=\frac{x}{55.5+x}\] |
| As \[55.5>>>x,\]thus neglecting from denominator |
| \[1{{0}^{-5}}=\frac{x}{55.5}\Rightarrow x=55.5\times 1{{0}^{-5}}\]Moles or \[~0.555\]millimoles. |
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