colligative properties

Question

If the boiling point of 400 g water has been increased by 9.3 C by adding lead (II) chloride, how many grams of lead (II) chloride were necessary to elicit this change? Assume the Kb of water is 0.51 C/m.

Answer 1

Lead (II) chloride = PbCl₂. This compound is not very soluble in water so this question is not as straight forward as it appears. To answer it properly, we would need the Ksp and works calculate the solubility first. If you want to assume it is completely soluble (an invalid assumption) then you would use

ΔT = imK

ΔT = change in b.p. = 9.3^o

i = vant Hoff factor = 3 for PbCl₂

m = molality = moles solute/kg solvent = ?

K = boiling constant = 0.51^o/m

Solve for m

m = ΔT / (i)(K) = 9.3 / (3)(0.59)

m = 5.25 moles / kg

Since we have only 0.400 kg (400 g) of water we can calculate moles of PbCl2 as

5.25 moles / kg x 0.400 kg = 2.10 moles

mass = 2.10 moles x 278 g/mole = 584 g PbCl2.

Answer 2

To solve this problem, we'll use the formula for the boiling point elevation:

ΔTb = i * Kf * m

Where:

- ΔTb is the boiling point elevation

- i is the van't Hoff factor (for lead (II) chloride, it's 2 since it dissociates into two ions in solution)

- Kf is the molal boiling point elevation constant of the solvent (water in this case)

- m is the molality of the solute (lead (II) chloride)

Given that ΔTb = 9.3°C, Kf = 0.51°C/m, and assuming complete dissociation of lead (II) chloride (i = 2), we can rearrange the formula to solve for molality (m):

m = ΔTb / (i * Kf)

Substituting the given values:

m = 9.3°C / (2 * 0.51°C/m)

m ≈ 9.12 m

Molality (m) is given by moles of solute per kilogram of solvent. Since we have 400 grams of water (which is equivalent to 0.4 kg), we can find the number of moles of lead (II) chloride necessary:

moles of lead (II) chloride = m * mass of solvent (in kg)

moles of lead (II) chloride = 9.12 m * 0.4 kg

moles of lead (II) chloride ≈ 3.648 moles

Now, we know that 1 mole of lead (II) chloride has a molar mass of approximately 278.1 g/mol (207.2 g/mol for lead + 2 * 35.45 g/mol for chlorine). Thus:

mass of lead (II) chloride = moles of lead (II) chloride * molar mass of lead (II) chloride

mass of lead (II) chloride ≈ 3.648 moles * 278.1 g/mol

mass of lead (II) chloride ≈ 1014.2 grams

So, approximately 1014.2 grams of lead (II) chloride were necessary to increase the boiling point of 400 grams of water by 9.3°C.