Q.

Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol. wt. of urea = 60).

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0.2 m, 0.00357

0.4 m, 0.00357

0.5 m, 0.00357

0.7 m, 0.00357

answer is A.

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Detailed Solution

Weight of solute (urea) dissolved = 3.0 gWeight of the solvent (water) = 250 gMolecular weight of the solute = 603.0 gm of the solute = 3.060 moles = 0.05 moles Molality = Number of moles of solute x 1000 g of solvent Weight of solvent∴ Molality =3/60250×1000=0.2mCalculation of mole fraction 3.0 g of solute = 3/60 moles = 0.05 mole 250 g of water =25018 moles = 13.94 moles ∴ Mole fraction of the solute =0.050.05+13.94=0.0513.99=0.00357

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