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

Options

A. 0.2 m, 0.00357
B. 0.4 m, 0.00357
C. 0.5 m, 0.00357
D. 0.7m, 0.00357

Answer

A

Explanation

Wt. of solute (urea) dissolved = 3.0 gm
Wt. of the solvent (water) = 250 gm
Mol. wt. of the solute = 60
3.0 gm of the solute = `3.0/60` moles = 0.05 moles
Thus 250 gm of the solvent contain = 0.05 moles of solute
`:.` 1000 gm of the solvent contain = `(0.05xx1000)/250=0.2` moles
Hence molality of the solution = 0.2 m
In short,
Molality = No. of moles of solute/1000 g of solvent
`:. "Molarity"=(3//60)/250xx1000=0.2 m`
Calculation of mole fraction :
3.0 gm of solute = 3/60 moles = 0.05 moles
250 gm of water = `250/18` moles = 13.94 moles
`:.` Mole fraction of the solute
`=0.05/(0.05+13.94)=0.05/13.99=0.00357`