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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    Complete combustion of 0.858 g of compound\[X\]gives 2.63 g of\[C{{O}_{2}}\]and 1.28 g of\[{{H}_{2}}O\]. The lowest molecular weight which\[X\]can have, is:

    A)  43 g                      

    B)         86 g

    C)  129 g                    

    D)         172 g

    E)  22 g

    Correct Answer: B

    Solution :

    \[%\]of \[C=\frac{12}{44}\times \frac{2.63}{0.858}\times 100=83.5%\] \[%\]if\[H=\frac{2}{18}\times \frac{1.28}{0.858}\times 100=16.5%\]
    Element % Relative no. of atoms Simplest ration
    C 83.5 83.5/12=7 7/7=1
    H 16.5 16.5/1=16.5 16.5/1 =2.35
                    \[C:H=1:2.35\] Multiply by 6 to make whole number\[6:14\]. Hence, empirical formula\[={{C}_{6}}{{H}_{14}}\] The minimum value of\[n=1\] Hence, its molecular formula is\[{{C}_{6}}{{H}_{14}}\]and molecular weight\[=86\text{ }g\].


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