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Two liquids A and B from an ideal solution. At 300K, the vapour pressure of a solution containing 1 mole of A and 3 moles of b is 550mmHg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of the solution increases by 10mmHg. Determine the vapour pressure of A and B in their pure states. |
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Answer» Solution :Mole FRACTION of A `=(1)/(1+3)=(1)/(4)` Mole fraction of B `=(3)/(1+3)=(3)/(4)` Applying EQN. 3 we get, `550=(1)/(4)xxp_(A)^(0)+(3)/(4)p_(B)^(0)`…………..(1) When ONE more mole of B was ADDED to the solution, Mole fraction of A `=(1)/(1+4)=(1)/(5)` Mole fraction of B `=(4)/(1+4)=(4)/(5)` Again applying Eqn. (3) , we get `550+10=(1)/(5)p_(A)^(0)+(4)/(5)p_(B)^(0)` ................(2) From (1) and (2), we get `p_(A)^(0)=440mm` `p_(B)^(0)=600mm` |
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