A) 90.92 mm Hg
B) 115.0 mm Hg
C) 147.9 mm Hg
D) 285.5 mm Hg
Correct Answer: A
Solution :
[a] \[^{n}CHC{{l}_{3}}=\frac{25.5}{119.5}=0.213\] \[^{n}C{{H}_{2}}C{{l}_{2}}=\frac{40}{85}=0.47\] \[{{P}_{T}}=P_{A}^{0}{{X}_{A}}+P_{B}^{0}{{X}_{B}}\] \[=200\times \frac{0.213}{0.683}+41.5\times \frac{0.47}{0.683}\] \[=62.37+28.55=90.92\]You need to login to perform this action.
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