2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    A solution containing 0.85 g of \[znC{{l}_{2}}\] in 125.0 g of water freezes at \[-0.23{}^\circ C\]. The apparent degree of dissociation of the salt is (\[{{K}_{f}}\] for water = 1.86 K kg \[\text{mo}{{\text{l}}^{-1}}\], atomic mass: Zn = 65.3 and Cl = 35.5)   JEE Main Online Paper (Held On 12 May 2012)

    A) 1.36%                                   

    B) 73.5%

    C) 7.35%                                   

    D) 2.47%

    Correct Answer: B

    Solution :

    Mol. wt.\[=\frac{{{k}_{f}}\times w\times 1000}{\Delta {{T}_{f}}\times W}=\frac{1.86\times 0.85\times 1000}{0.23\times 125}\approx 55\,gm\] Where w = 0.85g W = 125g \[\Delta {{T}_{f}}={{0}^{o}}C-(-{{23}^{o}}C)={{23}^{o}}C\] Now,\[i=\frac{{{M}_{normal}}}{{{M}_{observed}}}=\frac{136.3}{55}=2.47\] \[ZnC{{l}_{2}}{{\underset{\alpha }{\mathop{Zn}}\,}^{++}}+2\underset{2\alpha }{\mathop{C{{l}^{-}}}}\,\] Van't Hoff factor (0 \[=\frac{1-{{\alpha }_{\eta }}+{{\alpha }_{\eta }}+2{{\alpha }_{\eta }}}{1}=2.47\] \[\therefore \]\[{{\alpha }_{\eta }}=0.735=73.5%\]


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