Question

A solution of urea (Molar mass = 60 g mol–1) has been prepared by dissolving 6 g of urea in 500 g of water. The depression in freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1)

Answer 1

Molar mass of urea = 60 g

Mass of urea = 6 g

moles of urea = mass / molar mass

moles of urea = 6 /60

moles of urea =0.1

mass of water = 500 g

K f ( cryoscopic constant ) = 1.86 kg mol ^ -1

Depression in freezing point is given by the formula ,

∆ T f = K f × m

here ,

• ∆ T f =Depression in freezing point
• K f = cryoscopic constant
• m = molality

molality

= moles of solute / mass of solvent (Kg )

= moles of urea × 1000 / mass of water (g)

= 0.1 ×1000 / 500

= 1000/5000

= 0.2

molality ( m) = 0.2

∆ T f = K f × m

∆ T f = 1.86 × 0.2

∆ T f = 0.372