Question

17) Methanol, CH3OH, is sometimes used as an antifreeze for the water in automobile windshield washer fluids. How many moles of methanol must be added to 6.50 kg of water to lower its freezing point to -13.0 ∘C? For each mole of solute, the freezing point of 1 kg of water is lowered 1.86 ∘C.

18.) The following table gives the solubility of solute X in water at two different temperatures:

You have prepared a saturated solution of X at 20∘C using 33.0 g of water. How much more solute can be dissolved if the temperature is increased to 30∘C?

Temperature (∘C)	Solubility (g/100 g H2O)
20	11.0
30	23.0

16.) What is the freezing point of a solution that contains 25.4 g of urea, CO(NH2)2, in 205 mL water, H2O? Assume a density of water of 1.00 g/mL.

Express your answer to three significant figures and include the appropriate units.

13.) Which of the following solutions will have the highest boiling point?

0.50 mol HF in 2.0 kg of water

0.40 mol NaCl in 1.0 kg water

0.50 mol C6H12O6 in 1.0 kg of water

0.40 mol AlCl3 in 1.5 kg water

Answer 1

Q17) ∆T_f = i×K_f×m , where K_f is molal depression in freezing point, m is molality of solution , i is vant's Hoff factor = 1 for non dissociative solute and ∆T_f = freezing point of pure solvent - freezing point of solution.

0℃- (-13.0℃) = 13.0℃ = 1×1.86℃/m × (moles of solute/mass of solvent(in kg)) = 1.86 × (moles of CH₃OH/6.50) = 13.0

Moles of CH₃OH added = 13.0×6.50/1.86 = 45.43mole = 45.4mol. (Answer)

Q18:) amount of solute dissolved in 33.0ml water = (11.0/100)g × 33.0 = 3.63g at 20℃

Amount of solute dissolved in 33.0 ml water = (23.0/100)g×33.0 = 7.59g. at 30℃

Hence amount of more solute added = 7.59g - 3.63g

= 3.96g (answer)

Q16:)

Molality of urea solution = mass of urea/ molar mass × (1000/mass of solvent(in gm)) = 25.4g/60.06g/mol (1000/205)Kg = 2.06298molal

∆T_f = freezing point of solvent - freezing point of solution = K_f × m = 1.86℃/molal × 2.06298molal = 3.837℃

Freezing point of solution = 0℃ - 3.837℃ = - 3.837℃ = - 3.84℃. (Answer)

Q13:)

Answer : 0.40mol AlCl₃ in 1.5Kg water .

∆T_b = i×m×K_b

For dissociative electrolyte value of i is equal to number of ions formed from 1 molecule of solute.

i = 4 for AlCl₃

i=2 for NaCl

So effective number of ions will be more in AlCl₃ solution per Kg of water. Hence elevation in boiling point will maximum and so boiling point will also be maximum for this solution.