2. Solved Papers
  3. CET Karnataka Engineering

CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

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

    A) 2.292 kJ               

    B) 1.292kJ

    C) 0.292 kJ                               

    D) 3.392 kJ

    Correct Answer: A

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner