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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The amount of heat evolved when \[500\text{ }c{{m}^{3}}\] of \[0.1M\text{ }HCl\]is mixed with \[200\text{ }c{{m}^{3}}\]of \[0.2M\] \[NaOH\] is

    A)  \[2.292kJ\]

    B)  \[1.292kJ\]

    C)  \[0.292\text{ }kJ\]

    D)  \[3.392\text{ }kJ\]

    Correct Answer: A

    Solution :

    \[\underset{Initial\,moles}{\mathop{HCl}}\,+NaOH\to NaCl+{{H}_{2}}O\] \[\frac{500\times 0.1}{1000}\] \[\frac{200\times 0.2}{1000}\] \[0\] \[0\] \[=0.05\] \[=0.04\] Final moles             \[0.05-0.04\] \[0\] \[0.04\] \[0.04\] \[=0.01\]              . \[\because \] In neutralisation of 1 mole of \[NaOH\] by 1 mole \[HCl\], heat evolved \[=57.3\text{ }kJ\] \[\therefore \]To neutralise \[0.04mole\]of \[NaOH\] by \[0.04mole\]of \[HCl\], heat evolved will be \[=57.3\times 0.04\text{ }kJ\] \[=2.292kJ\]


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