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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)
Solutions
Solution:
Wt. of solute (urea) dissolved = 3.0 g
Wt. of the solvent (water) = 250 g
Mol. wt. of the solute = 60
3.0 gm of the solue $=\frac{3.0}{60}$ moles
$=0.05$ moles
Thus 250 g of the solvent contain = 0.05 moles of solute
$\therefore $ 1000 g of the solvent contain
$=\frac{0.05\times 1000}{250}=0.2$ moles
Molality = No. of moles of solute/1000 g of solvent
$\therefore $ Molality $=\frac{3/60}{250}\times 1000=0.2\,m$
Calculation of mole fraction
3.0 g of solute =3/60 moles =0.05 mole
250 g of water $=\frac{250}{18}$ moles $=13.94$ moles
$\therefore $ Mole fraction of the solute
$=\frac{0.05}{0.05+13.94} = \frac{0.05}{13.99}$
$=0.00357$