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NEET Sample Paper NEET Sample Test Paper-80

  • question_answer
    Addition of 0.643 g of a compound to 50 mL of benzene\[\left( density = 0.879 g m{{L}^{-}}^{1} \right)\], lowers the freezing point from \[50.51{}^\circ C \,to \,50.03{}^\circ C\]. If \[{{K}_{f}}\] for benzene is 5.12, the molecular mass of the compound is

    A)  156.00            

    B)  312.00

    C)  78.00              

    D)  468.00

    Correct Answer: A

    Solution :

    Weight of benzene \[=V\,\times  d = 50 \times  0.879 g\] \[=43.95 g\] \[\operatorname{Weight} of compound = 0.643 g\] \[\Delta {{T}_{f}}\,\,=\,\,50.51-50.03=0.48\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{T}_{f}}\,\,=\,\,{{K}_{f}}\times \,\,\frac{w}{M}\,\,\times \,\,\frac{1000}{W}\] \[0.48=5.12\times \frac{0.643}{M}\times \frac{1000}{43.95}\] \[\operatorname{M}\,\,=\,\,156.05 g\,mo{{l}^{-1}}\]


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