A) \[158\]
B) \[257\]
C) \[137\]
D) \[197\]
Correct Answer: A
Solution :
\[P={{x}^{1}}{{P}^{o}}_{C{{H}_{3}}CHBr}+{{x}^{2}}{{P}^{o}}_{C{{H}_{2}}BrC{{H}_{2}}Br}\] For \[C{{H}_{3}}CHBr\] \[{{x}^{1}}=\frac{2}{2+1}=\frac{2}{3}\] For \[C{{H}_{2}}BrC{{H}_{2}}Br\] \[{{x}^{2}}=\frac{1}{2+1}=\frac{1}{3}\] \[P=\frac{2}{3}\times 173+\frac{1}{3}\times 127\] \[=157.66=158\]You need to login to perform this action.
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