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Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions
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Q)

The vapour pressure of pure liquid solvent A is 0.80atm.When a non volatile substance B is added to the solvent,its vapour pressure drops to 0.40atm,the mole fraction of component B in the solution is

$\begin{array}{1 1}(a)\;0.25\\(b)\;0.75\\(c)\;0.50\\(d)\;0.40\end{array}$

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