Question

(Show your work please)

Answer 1

`\boxed{\text{bp =100.106 }^(\circ)\text{C; fp = -0.387 }^(\circ)\text{C; bp =101.68 }^(\circ)\text{C}}`

Step-by-step explanation:

Q8. Boiling point

Data:

m(KOH) = 53.1 g

m(H₂O) = 9.10 kg

K_b = 0.512 °C·kg·mol⁻¹

Calculations:

(a) Moles of KOH

`\text{Moles of KOH} = \text{53.1 g KOH} * \frac{\text{1 mol KOH}}{\text{56.11 g KOH}} = \text{0.9464 mol KOH}`

(b) Molal concentration

The formula for molal concentration (b) is

`b = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = (0.9464)/(9.1) = \text{0.104 mol/kg}`

(c) Boiling point elevation

The formula for the boiling point elevation ΔTb is

`\Delta T_(b) = iK_(b)b`

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For KOH,

KOH(s) ⟶ K⁺(aq) + OH⁻(aq)

1 mol KOH ⟶ 2 mol particles i = 1

`\Delta T_(b) = 2 * 0.512 * 0.104 = \text{0.106 }^(\circ)\text{C}`

(d) Boiling point

`T_(b) = T_(b)^(\circ) + \Delta T_(b) = 100.000 + 0.106 = \text{100.106 }^(\circ)\text{C}`

Q9. Freezing point

`\Delta T_(f) = iK_(f)b = 2 * 1.86 * 0.104 = \text{0.387 }^(\circ)\text{C}\\\\T_(f) = T_(f)^(\circ) - \Delta T_(f) = 0.000 - 0.387 = \text{-0.387 }^(\circ)\text{C}`

Q10. Boiling point

`\text{Kilograms of phenol} = 645 * (1)/(1000) = \text{0.645 kg phenol}\\\\b = (0.910)/(0.645) = \text{1.411 mol/kg}\\\\\Delta T_(b) = 1* 1.19 * 1.411 = \text{1.68 }^(\circ)\text{C}\\T_(b) = 100.00 + 1.68 = \text{101.68 }^(\circ)\text{C}`