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AMU Medical AMU Solved Paper-2001

  • question_answer
    Percentage composition of an organic compound is as follows\[C=10.06,\,\,\,\,H=0.84,\,\,\,Cl=89.10\] Which of the following correspond to its molecular formula if the vapour density is 60.0?

    A)  \[C{{H}_{3}}Cl\]                              

    B)  \[CHC{{l}_{3}}\]

    C)  \[C{{H}_{2}}C{{l}_{2}}\]               

    D)  None of these

    Correct Answer: B

    Solution :

                     Key Idea Molecular formula \[=\text{ }{{\left( empirical\text{ }formula \right)}_{n}}\] where, \[n=\frac{molecular\text{ }formula\text{ }weight}{empirical\text{ }formula\text{ }weight}\] Molecular weight = 2 \[\times \] vapour density Vapour density = 60
    Element % Atomic weight \[\frac{a}{b}=x\] Ration
    C 10.06 12 \[\frac{10.06}{12}=0.838\] 1
    H 0.84 1 \[\frac{0.84}{1}=0.84\] 1
    \[Cl\] 89.10 35.5 \[\frac{89.10}{35.5}=2.50\] 3
    Empirical formula \[=CHC{{l}_{3}}\] Empirical formula mass = 119.5 Molecular weight \[=2\times 60=120\] \[n=\frac{molar\text{ }mass}{empirical\text{ }formula\text{ }mass}=\frac{120}{119.5}\] n = 1 \[\therefore \] Molecular formula \[={{(CHC{{l}_{3}})}_{1}}=CHC{{l}_{3}}\]


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