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If mole fraction of solvent is ' \( s \) ' and lowering of vapour pressure is ' \( p \) ' mm of Hg then, what will be vapour pressure of pure solvent (Po)? (assume solute is nonvolatile and nonelectrolyte) (1) \( p^{\circ}=s p \) (2) \( p ^{\circ}= p ^{\prime} s \) (3) \( p^{\circ}=s / p \) (4) \( p^{0}=p /(1-s) \) |
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Answer» Correct option is (4) Po = \(\frac{P}{1-S}\) As we know, relative lowering of vapour pressure is given by-- \(\frac{P^o-P_1}{P^o}=x\)----(1) where, Po - P1 = lowering in vapour pressure Po = Vapour pressure of pure solvent x = mole fraction of solute we have given, lowering of vapour pressure = Po - P1 = P mole fraction of solvent = 'S' ∴ mole fraction of solute = 1 - S putting these values in equation (1) \(\frac{P}{P^o}=1-S\) Po = \(\frac{P}{1-S}\) Po = \(\frac{P}{1-S}\) Therefore, vapour pressure of pure solvent will be Po = \(\frac{P}{1-S}\) |
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