CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    Heat liberated with \[100ml\]of \[1N\text{ }NaOH\]is neutralised by \[300ml\]of IN \[HCl\]:

    A)  \[11.56kJ\]

    B)  \[5.73kJ\]

    C)  \[22.92kJ\]

    D)  \[17.19kJ\]

    Correct Answer: B

    Solution :

    The molarity and normality are same in case of \[NaOH\] and \[HCl\] because the acidity and basicity of these are one and one respectively. Since, number of moles = molarity \[\times \]volume (I) \[\therefore \] No. of moles of \[NaOH\] solution \[=\frac{100\times 1}{1000}=0.1\] and number of moles of \[HCl\] solution \[=\frac{300\times 1}{1000}=0.3\] Hence, 0.1 mole \[NaOH\] is neutralised by 0.1 mole \[HCl\]. We know that heat of neutralisation of 1 mole\[HCl\] and 1 mole \[NaOH\] is 57.3 kJ. Hence, heat of neutralisation of 0.1 mol \[HCl\] and 0.1 mole \[NaOH\]will be \[=57.3\times 0.1=5.73kJ\]


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