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Question

The vapour pressure of two pure liquids A and B are 50 and 40 torr respectively. If 8 moles of A is mixed with x moles of B, then vapour pressure of solution obtained is 48 torr. What is the value of x?


Solution

We have 
The vapour pressure of two pure liquids $$A$$ and $$B$$ are $$50$$ and $$40$$ torr respectively.
Now, according to question 
We have to $$8mol$$ of $$A$$ is mixed with $$x$$ mole of $$B$$
Now, from the Raults law
$${P_T} = p_A^ \circ  + P_B^ \circ $$

$$ = {P_A} \times {x_A} + {P_B} \cdot \left( {1 - {x_A}} \right)$$

$$ = {P_A} \times {x_A} + {P_B} - {P_B} \times {x_A}$$

$$ = \left( {{P_A} - {P_B}} \right){x_A} + {P_B}$$
Now,
$$ \Rightarrow 48 = \left( {50 - 40} \right){x_A} + 40$$  [Given form the the question the vapour pressure of soution ]
Hence.
$${x_A} = 0.8$$
and,
$${x_B} = 0.2$$.

Chemistry

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