J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer
    Molality of 2.5 g of ethanoic acid \[(C{{H}_{3}}COOH)\] in 75 g of benzene is

    A) \[~0.565\text{ }mol\text{ }k{{g}^{-1}}\]     

    B) \[~0.656\text{ }mol\text{ }k{{g}^{-1}}\]

    C) \[~0.556\text{ }mol\text{ }k{{g}^{-1}}\]     

    D) \[0.665\text{ }mol\text{ }k{{g}^{-1}}\]

    Correct Answer: C

    Solution :

     Molality (m) \[\text{=}\,\frac{\text{mass}\,\text{of}\,\text{solute}\,\text{(in}\,\text{g)}\,\text{ }\!\!\times\!\!\text{ }\,\text{1000}}{\text{molecular}\,\text{weight}\,\text{of}\,\text{solute}\,\text{ }\!\!\times\!\!\text{ }\,\text{massofsolvent(in}\,\text{g)}}\] \[=\frac{2.5\times 1000}{60\times 7.5}=0.555\,\text{mol}\,\text{k}{{\text{g}}^{-1}}\]


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