A) 150kPa
B) 75 kPa
C) 100.0kPa
D) 72.5kPa
Correct Answer: D
Solution :
Molecular mass of heptane \[N{{O}_{2}}\], \[\therefore \] \[{{n}_{2}}=2{{n}_{1}}\] Molecular mass of octane \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{{{T}_{1}}}{2{{T}_{2}}}\] \[=\frac{273}{2(546)}\] \[\therefore \] \[{{p}_{2}}=\frac{2(546)}{273}{{p}_{1}}=4{{p}_{1}}\] and \[{{p}_{2}}\] Thus, \[\underset{2,2\,\,4-trimethyl\,pen\tan e}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\] \[{{H}_{3}}C-C\equiv CH\xrightarrow[1%{{H}_{2}}S{{O}_{4}}]{40%H{{ & }_{2}}SO{{ & }_{4}}}{{H}_{3}}C-\underset{\begin{smallmatrix} | \\ \underset{\begin{smallmatrix} prop-1en-2ol \\ (Unstable) \end{smallmatrix}}{\mathop{OH}}\, \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\] \[\xrightarrow{Tautomerism}{{H}_{3}}C\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{-C}}\,-C{{H}_{3}}\] \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{-CH}}\,-C{{H}_{3}}+alc.KOH\to \]
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