A) 173.9 mmHg
B) 615.0 mmHg
C) 347.9 mmHg
D) 28.5 mmHg
E) None of the above
Correct Answer: E
Solution :
Number of moles of \[CHC{{l}_{3}},\] \[{{n}_{A}}=\frac{w}{M}\] \[=\frac{25.5}{119.5}\] \[=0.213\] Number of moles of \[C{{H}_{2}}C{{l}_{2}},\] \[{{n}_{B}}=\frac{40}{85}=0.47\] Mole fraction of \[CHC{{l}_{3}}\] \[{{\chi }_{A}}=\frac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}\] \[=\frac{0.213}{0.683}=0.31\] Mole fraction of \[C{{H}_{2}}C{{l}_{2}},\] \[{{\chi }_{B}}=1-{{\chi }_{A}}\] \[=1-0.31=0.69\] \[\because \] \[{{p}_{total}}={{p}_{A}}{{\chi }_{A}}+{{p}_{B}}{{\chi }_{B}}\] \[=200\times 0.31+41.5\times 0.69\] \[=62+28.63\] \[=90.63mm\,Hg\]You need to login to perform this action.
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