Answer:
[tex]\boxed{\text{bp =100.106 }^{\circ}\text{C; fp = -0.387 }^{\circ}\text{C; bp =101.68 }^{\circ}\text{C}}[/tex]
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
[tex]\text{Moles of KOH} = \text{53.1 g KOH} \times \dfrac{\text{1 mol KOH}}{\text{56.11 g KOH}} = \text{0.9464 mol KOH}[/tex]
(b) Molal concentration
The formula for molal concentration (b) is
[tex]b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}} = \dfrac{0.9464}{9.1} = \text{0.104 mol/kg}[/tex]
(c) Boiling point elevation
The formula for the boiling point elevation ΔTb is
[tex]\Delta T_{b} = iK_{b}b[/tex]
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
[tex]\Delta T_{b} = 2 \times 0.512 \times 0.104 = \text{0.106 }^{\circ}\text{C}[/tex]
(d) Boiling point
[tex]T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.000 + 0.106 = \text{100.106 }^{\circ}\text{C}[/tex]
Q9. Freezing point
[tex]\Delta T_{f} = iK_{f}b = 2 \times 1.86 \times 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}[/tex]
Q10. Boiling point
[tex]\text{Kilograms of phenol} = 645 \times \dfrac{1}{1000} = \text{0.645 kg phenol}\\\\b = \dfrac{0.910}{0.645} = \text{1.411 mol/kg}\\\\\Delta T_{b} = 1\times 1.19 \times 1.411 = \text{1.68 }^{\circ}\text{C}\\T_{b} = 100.00 + 1.68 = \text{101.68 }^{\circ}\text{C}[/tex]
