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JEE Main & Advanced Chemistry Some Basic Concepts of Chemistry / रसायन की कुछ मूलभूत अवधारणाएँ Question Bank Mock Test - Some Basic Concepts of Chemistry

  • question_answer
    The minimum quantity in gram of \[{{H}_{2}}S\] needed to precipitate 63.5 g of \[C{{u}^{{{2}^{+}}}}\] will be \[(C{{u}^{{{2}^{+}}}}+{{H}_{2}}S\to C{{u}_{2}}S+{{H}_{2}})\] Black

    A) 63.5 g  

    B) 31.75 g

    C) 34 g     

    D) 2 g

    Correct Answer: C

    Solution :

    \[C{{u}^{2+}}+{{H}_{2}}S\to C{{u}_{2}}S+{{H}_{2}}\] 63.5g 1 mole of \[C{{u}^{2+}}\]requires 1 mole of \[{{H}_{2}}S\] Or 63.5 g of \[C{{u}^{2+}}\]requires 34 g of \[{{H}_{2}}S\] [Molar mass of \[{{H}_{2}}S\]=2+32=34g] So 1 mole \[{{H}_{2}}S\]is required, i.e, 34 g      


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