Question

Calculate the boiling point and freezing point of a 20% aqueous sucrose C12H22O11 (MM=342 solution

Answer 1

Answer on Question #49788 - Chemistry - Other

Question

Calculate the boiling point and freezing point of a $20\%$ aqueous sucrose $C_{12}H_{22}O_{11}$ (MM=342) solution.

Answer:

The boiling point elevation is proportional to the molality of the solute particles:

$$\Delta T _ {\mathrm {b}} = K _ {\mathrm {b}} m$$

$\Delta T_{\mathrm{b}} =$ the amount by which the boiling point is raised, $m =$ molality (moles solute particles per kg of solution), $K_{\mathrm{b}} =$ molal boiling-point elevation constant (solvent dependent), for water $K_{\mathrm{b}} = 0.512^{\circ}\mathrm{Cm}^{-1}$. Boiling Point of solution = normal boiling point of solvent + $\Delta T_{\mathrm{b}}$. The freezing point depression is proportional to the molality of the solute particles:

$$\Delta T _ {f} = - K _ {f} m$$

$\Delta T_{\mathrm{f}} =$ the amount by which the freezing point is lowered, $m =$ molality (moles solute particles per kg of solution), $K_{\mathrm{f}} =$ molal freezing-point depression constant (solvent dependent), for water $K_{\mathrm{f}} = 1.855^{\circ}\mathrm{Cm}^{-1}$

Freezing Point of solution = normal freezing point of solvent + $\Delta T_{\mathrm{f}}$

Assume we have $100\mathrm{g}$ of $20\%$ aqueous sucrose $C_{12}H_{22}O_{11}$ solution, then the mass of sucrose is $20\mathrm{g}$ and the mass of water is $80\mathrm{g}$. Molality of this solution is:

$$m = \frac {m (\text {sucrose})}{M (\text {sucrose}) m (\text {water})} = \frac {20}{342 \cdot 80} = 0.000731 \mathrm{mol}/\mathrm{g} = 0.731 \mathrm{mol}/\mathrm{kg}$$

The boiling point elevation is:

$$\Delta T _ {b} = 0.512 \cdot 0.731 = 0.374$$

Boiling Point of solution = normal boiling point of solvent + $\Delta T_{\mathrm{b}} = 100.0 + 0.374 = 100.374$ °C

The freezing point depression is:

$$\Delta T _ {f} = -1.855 \cdot 0.731 = -1.356$$

Freezing Point of solution = normal freezing point of solvent + $\Delta T_{\mathrm{f}} = 0.0 + (-1.356) = -1.356$ °C

Answer: $T_{\mathrm{b}} = 100.374^{\circ} \mathrm{C}, T_{\mathrm{f}} = -1.356^{\circ} \mathrm{C}$.

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