An X molal solution of a compound in benzene has mole fraction of solute $$=$$ $$0.2$$. The value of X is

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Infinity Learn · Accepted Answer

Mole fraction of solute $$X2=0.2$$. Therefore, mole fraction of solvent $$X1=0.8Or$$ $$n2n1+n2=0.2$$ and $$n1n1+n2=0.8Or$$ $$n2n1=0.20.8=14Now$$, if $$n1$$ (solvent$$ $$moles) $$=$$ $$1000/78$$ $$=$$ $$12.8$$ moles $$n2=12.8/4=3.2$$ moles. Therefore, $$3.2$$ moles of the compound are present in one kg of solvent benzene and so molality $$=$$ $$3.2$$.                                                                                   ORMole fraction of solute $$=$$  $$mm+1000Molecular$$ weight of solvent                           $$0.2$$   $$=$$ $$mm+100078$$                     ⇒$$0.2$$ $$=$$ $$mm+12.82$$                       m $$=$$ $$0.2m+$$ $$12.82$$×$$0.2$$                       m $$=$$ $$3.2$$

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