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NEET Sample Paper NEET Sample Test Paper-70

  • question_answer
    A solution containing 30 g of non-volatile solute in exactly 90 g water has a vapour pressure of 21.85 mm Hg at \[25{}^\circ C\]. Further 18 g of waters then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at\[25{}^\circ C\]. Calculate the molecular weight of the solute.

    A) 74.2                

    B) 75.6

    C) 67.83              

    D) 78.7              

    Correct Answer: C

    Solution :

    We have \[\frac{{{p}^{0}}-21.85}{21.85}=\frac{30\times 18}{90\times m},\,\,\]               for I case (i) Weight of solvent \[= 90 + 18 = 108 g\] \[\frac{{{p}^{0}}-22.15}{22.15}=\frac{30\times 18}{108\times m},\]                 for II case (ii) By eq.[A] \[{{p}^{0}}_{m}\,=\,\,21.85m\,\,=\,\,21.85\,\,\times \,\,6=131.1\] By eq. [B] \[p_{m}^{0}-22.15\,m\, =\, 22.15 \times  5 = 110.75\] \[0.30\,m\,\,=\,\,20.35\] \[m=\frac{20.35}{0.30}=67.83\]


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