Question

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)

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

Answer 1

Answer: (A)

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 $=\frac{3/60}{250}\times 1000=0.2m$
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
$=\frac{0.05}{0.05+13.94}=\frac{0.05}{13.99}$
$=0.00357$