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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    Equal masses of a solute are dissolved in equal amount of two solvents A and B, respective molecular masses being \[{{M}_{A}}\] and \[{{M}_{B}}\]. The relative lowering of vapour pressure of solution in solvent A is twice that of the solution in solvent B. If the solutions are dilute, \[{{M}_{A}}\] and \[{{M}_{B}}\] are related as

    A) \[{{M}_{A}}={{M}_{B}}\]                  

    B) \[2{{M}_{A}}={{M}_{B}}\]

    C) \[{{M}_{A}}=2{{M}_{B}}\]                

    D) \[{{M}_{A}}=4{{M}_{B}}\]

    Correct Answer: C

    Solution :

    [c] For dilute solution, \[\frac{\Delta P}{P{}^\circ }\Rightarrow \frac{{{n}_{solute}}}{{{n}_{solvent}}}\] For solution in A, \[\frac{\Delta {{P}_{A}}}{P_{A}^{{}^\circ }}=\frac{W/M}{{{W}_{A}}/{{M}_{A}}}=\frac{W}{M}\times \frac{{{M}_{A}}}{{{W}_{A}}}\]         ...(i) For solution in B, \[\frac{\Delta {{P}_{B}}}{P_{A}^{{}^\circ }}=\frac{W}{M}\times \frac{{{M}_{B}}}{{{W}_{B}}}\]   ....(ii) From (i) and (ii), \[\frac{\Delta {{P}_{A}}/{{P}^{{}^\circ }}_{A}}{\Delta {{P}_{B}}/{{P}^{{}^\circ }}_{B}}=2=\frac{{{M}_{A}}{{W}_{B}}}{{{M}_{B}}{{W}_{A}}}=\frac{{{M}_{A}}}{{{W}_{B}}}({{W}_{A}}={{W}_{B}})\] \[{{M}_{A}}=2{{M}_{B}}\]


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