2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 8

  • question_answer
    One mole of \[NaCl\] (s) on melting absorbed \[30.5\text{ }kJ\]one of heat and its entropy is increased by \[28.8\,J{{K}^{-1}}mo{{l}^{-1}}\]. The melting point of \[NaCl\] is

    A) \[1059K\]          

    B) \[30.5K\]   

    C) \[28.8\,K\]         

    D) \[28800\,K\]

    Correct Answer: A

    Solution :

    \[NaCl(s)\to NaCl(l)\] Given that: \[\Delta H=30.5\,kJ\,mo{{l}^{-1}}\] \[\Delta S=28.8\,J{{K}^{-1}}mo{{l}^{-1}}=28.8\times {{10}^{-3}}kJ\,{{K}^{-1}}mo{{l}^{-1}}\] By using \[\Delta S=\frac{\Delta H}{T}=\frac{30.5}{28.8\times {{10}^{-3}}}=1059\,K.\]


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