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Manipal Medical Manipal Medical Solved Paper-2003

  • question_answer
    According to equation, \[{{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}2{{H}_{2}}O(l)+6C{{O}_{2}}(g)\] \[\Delta H=-3264.4\text{ }kJ/mol\]. The energy evolved when 7.8 g of benzene is burnt in air will be:

    A)  3.264 kJ/mol  

    B)  32.64 kJ/mol

    C)  326.4 kJ/mol  

    D)  163.22 kJ/mol

    Correct Answer: C

    Solution :

     \[\underset{(6\times 12+6\times 1=78)}{\mathop{{{C}_{6}}{{H}_{6}}(l)+}}\,\frac{15}{2}{{O}_{2}}(g)\to 3{{H}_{2}}O(l)+6C{{O}_{2}}(g)\] \[\because \]The energy evolved when 78 g. benzene is burnt = 3264.4 kJ/mol \[\therefore \]The energy evolved when 7.8g benzene is burnt \[=\frac{3264.4\times 7.8}{78}=326.44\text{ }kJ/mol\]


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