The vapour pressure of a `5%` aqueous solution of a non-volatile organic substance at `373 K`. Is `745 mm`. Calculate the molecular mass of the solute.

The vapour pressure of a `5%` aqueous solution of a non-volatile organ

1.	The vapour pressure of a `5%` aqueous solution of a non-volatile organic substance at `373 K`. Is `745 mm`. Calculate the molecular mass of the solute.
Answer» `5%` aqueous solution of the solute implies that `5 g` of the solute are present in `100 g` of the solution, i.e., Weight of solute `(W)_(2)=5 g` Weight of solution=100 g `:.` Weight of solvent `(W_(1))=100-5 =95 g` Further as the solution is aqueous, it means that the solvent is water and we know that vapour pressure of pure water at `393 K =760mm` Vapour pressure of solvent at `373 K (P_(s))=745 mm` (given) Molecular mass of solvent (water), `Mw_(1)=18` Molecular massof solute, `Mw_(2)`= To be calculated Using the formula for dilute solutions, viz `(P^(@)-P_(s))/P^(@)=n_(2)/n_(1)=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))` We get (760-745)/760=(5//Mw_(2))/(95//18) ` or `Mw_(2)=(5 xx 18 xx 760)/(15 xx 95)=48`

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The vapour pressure of a `5%` aqueous solution of a non-volatile organic substance at `373 K`. Is `745 mm`. Calculate the molecular mass of the solute.

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