Haryana PMT Haryana PMT Solved Paper-2007

  • question_answer
    \[Ag\] crystallises as fee. If radius of \[Ag\] is \[~144\text{ }pm\]then its density will be        

    A)  \[10ge{{m}^{-3}}\]        

    B)  \[5\text{ }ge{{m}^{-3}}\]

    C)  \[15\text{ }ge{{m}^{-3}}\]          

    D)  \[6.5\text{ }ge{{m}^{-3}}\]

    Correct Answer: A

    Solution :

                    For fee, radius \[=\frac{\sqrt{2}}{4}a=0.3535a\] Radius of \[Ag(r)=144\times {{10}^{-10}}cm,\] \[a=\frac{144\times {{10}^{-10}}}{0.3535}=407.35\times {{10}^{-10}}\] Density \[(\rho )=\frac{Z\times M}{{{a}^{3}}\times {{N}_{0}}}\] \[=\frac{4\times 108}{{{(407.35\times {{10}^{-10}})}^{3}}\times 6.02\times {{10}^{23}}}\] \[=\frac{432}{6.76\times {{10}^{-23}}\times 6.02\times {{10}^{23}}}\] \[=\frac{432}{40.7}=10.6=10g\,c{{m}^{-3}}\]


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