2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    A solution containing 62 g ethylene glycol in 250 g water is cooled to\[-10{}^\circ C\]. If \[{{K}_{f}}\] for water is \[1.86\text{ }K\text{ }kg\text{ }mo{{l}^{-1}}\], the amount of water (in g) separated as ice is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 32                    

    B)   16

    C) 64         

    D)   48

    Correct Answer: C

    Solution :

                \[\Delta \,{{T}_{f}}\,=\,{{K}_{f}}\,\times \,m\] \[10\,\,=\,1.86\,\times \,\frac{62/62}{{{W}_{(gm)}}}\,\times \,1000\] \[W=186\,gm\] \[\therefore \] Amount of water separated as ice \[=\left( 250-186 \right)=64\,gm\]


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