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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2014

  • question_answer
    \[3\Omega \]of gelatin is required to be added \[4\Omega \]of a standard gold solution to just prevent its precipitation by the addition \[4.5\Omega \]of 10% \[NaCl\] solution to it. Hence, the gold number of gelatin in milligram is 

    A) \[5\Omega \]                    

    B) \[\frac{\sqrt{3}}{1}\]            

    C) \[\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\]                                        

    D) \[\frac{(\sqrt{3}+1)}{1}\]  

    Correct Answer: B

    Solution :

                    \[M{{L}^{2}}\]   \[N{{(C{{H}_{3}})}_{2}}H>N(C{{H}_{3}}){{H}_{2}}>N{{(C{{H}_{3}})}_{3}}>N{{H}_{3}}\] \[NaCl\]                solution prevent coagulation when gelatin added \[O{{H}^{-}}\] \[{{H}_{2}}O\]   \[\text{C}{{\text{u}}_{\text{2}}}\] \[{{\left( \text{SCN} \right)}_{\text{2}}}.\] solution prevent coagulation when gelatin added \[{{K}_{6}}=\frac{RT_{0}^{2}}{1000\Delta {{H}_{v}}}\] So, gold number of gelatin in milligram \[=\frac{0.002\times {{(373.15)}^{2}}}{1000\times \left( \frac{9.72}{18} \right)}\]


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