Question

a.) Estimate the freezing point of 250 mL of water to which 8.5g of magnesium nitrate has been added. treat each solution as ideal.

b.) If we were to use 8.5g of strontium nitrate instead of magnesium nitrate, would the freezing temperature be higher or lower ( or unchanged) compared to solution in part a?

c.) if we were to use 8.5g of aluminum nitrate instead of magnesium nitrate ( as in part a), would the freezing point be hight or lower ( or unchanged) compared to solution in part a?

Answer 1

Mg(NO3)2 moles = mass / molar mass = 8.5 /148.3 = 0.057316

i = vantoff factor = 3 for MgN(NO3)2 since it gives 3 ions Mg2+ and 2NO3- as Mg(NO3)2 completely dissociates

water mass = 250 g ( as water density = 1g/ml , mass = vol x density)

molality - moles of solute / solvent mass in kg = 0.057316 /0.25 = 0.22926

now we have formula dT = i x Kf x m where Kf = 1.86 for water

hence dT = 3 x 1.86 x 0.22926

dT = 1.28 = Freezing point of water - Freezing point of solution      ( FP of water = 0C)

Freezing point of solution = 0 - 1.28 = - 1.28 C

b) Sr(NO3)2    moles = 8.5 /211.63 = 0.04

hence due to higher molecular mass of Strontium nitrate , we have lessr moles , hence lessoer molality and hence we have Higher freezing point compared to   (a)

C) Aluminium nitrate Al( NO3)3   gives 4 iosn , hence i = 4 , moles of AL(NO3)3 = 8.5/213 =0.0399

i x moles = 4 x 0.0399 = 0.1596 which is less than Mg(NO3)3 ( i x moles for it is 0.057316 x 3 = 0.172)

hence we get higher freezing point for this solution compared to part a