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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of a pure liquid A is 80 mmHg at 300K.It forms an ideal solution with liquid B.When the mole fraction of B is 0.4,the total pressure was found to be 88 mmHg.The vapour pressure of liquid A would be

$\begin{array}{1 1}\text{84 mmHg}\\\text{90 mmHg}\\\text{100 mmHg}\\\text{120 mmHg}\end{array} $

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Answer : 100 mmHg
$x_Ap_A^*+x_Bp_B^*=p$
$\Rightarrow 0.6\times 80$ mmHg+$0.4p_B^*=88$ mmHg
Hence $p_B^*=\large\frac{(88-0.6\times 80)mmHg}{0.4}$
$\Rightarrow $100 mmHg
answered Jul 22, 2014 by sreemathi.v
 

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