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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Mock Test - Solutions

  • question_answer
    The elevation in boiling point of a solution of 13.44 g of\[CuC{{l}_{2}}\], (molecular weight = 134.4, \[{{K}_{b}}\] = 0.52 K \[molalit{{y}^{-1}}\] in 1 kg water using the given information will be

    A) 0.16                 

    B) 0.05

    C) 0.1                   

    D) 0.2

    Correct Answer: A

    Solution :

    [a]
    \[CuC{{l}_{2}}\to C{{u}^{2+}}+2C{{I}^{-}}\]
    Initial 1 0 0
    Final                   \[(1-\alpha )\] \[\alpha \] \[2\alpha \]
    As we know that \[i=\frac{No.\text{ }of\text{ }particles\text{ }after\text{ }ionisation}{No.\text{ }of\text{ }particles\text{ }before\text{ }ionisation}\] \[\Delta {{T}_{b}}=i\times {{k}_{b}}\times m\] \[\therefore i=\frac{1+2\alpha }{1}\Rightarrow i=1+2\alpha \] Assuming 100% ionization \[i=1+2=3\] Molality \[=\frac{{{w}_{solute}}}{{{M}_{solute}}+{{M}_{solvent(KG)}}}=\frac{13.44}{134.4\times 1}=0.1\] \[\therefore \Delta {{T}_{b}}=3\times 0.52\times 0.1=0.156\approx 0.16\]


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