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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    \[KCl\]crystallises in the same type of lattice as does\[NaCl\].  Given that\[{{r}_{Na}}+/{{r}_{C{{l}^{-}}}}=0.55\]and\[{{r}_{{{K}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.74\]. Calculate the ratio of the side of the unit cell for\[KCl\]to that of\[NaCl\].

    A) 1.123                                    

    B) 0.0891

    C) 1.414                    

    D)        0.414

    E) 1.732

    Correct Answer: A

    Solution :

    Given      \[{{r}_{N{{a}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.55\] \[{{r}_{{{K}^{+}}}}/{{r}_{C{{l}^{-}}}}=0.74\] \[\frac{{{r}_{KCl}}}{{{r}_{NaCl}}}=?\] \[\frac{{{r}_{N{{a}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.55\] \[\frac{{{r}_{N{{a}^{+}}}}}{{{r}_{C{{l}^{-}}}}}+1=0.55+1\] \[\frac{r_{Na}^{+}+{{r}_{C{{l}^{-}}}}}{{{r}_{C{{l}^{-}}}}}=1.55\]                    ?? (i)                 \[\frac{{{r}_{{{K}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.74\]                 \[\frac{{{r}_{{{K}^{+}}}}}{{{r}_{C{{l}^{-}}}}}+1=0.74+1\]                 \[\frac{{{r}_{{{K}^{+}}}}+{{r}_{C{{l}^{-}}}}}{{{r}_{C{{l}^{-}}}}}=1.74\]                          ?.. (ii) Eqs (ii)/(i)                 \[\frac{{{r}_{{{K}^{+}}}}+{{r}_{C{{l}^{-}}}}}{{{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}}}=\frac{1.74}{1.55}=1.1226\]


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