Question

Calculate the freezing point and boiling point of each of the following solutions:
1. the freezing point of the solution: 222 g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 1.40 kg of water (Kf=1.86∘C)
Express your answer using one decimal place.

2.the boiling point of the solution: 222 g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 1.40 kg of water (Kb=0.52∘C)
Express your answer using one decimal place.

Answer 1

Answer on Question #56062 - Chemistry - General chemistry

Question:

Calculate the freezing point and boiling point of each of the following solutions:

1. the freezing point of the solution: 222 g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 1.40 kg of water (Kf=1.86°C)

Express your answer using one decimal place.

2. the boiling point of the solution: 222 g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 1.40 kg of water (Kb=0.52°C)

Express your answer using one decimal place.

Answer:

1. The change of the freezing point can be found using the following equation:

$$\Delta t = K_f \times C$$

where $K_f$ – the cryoscopic constant ($K_f = 1.86 \, \text{K kg mol}^{-1}$ for water) and $C$ – the molality of the solution.

$C = v / M$

where $v$ – the number of moles of dissolved compound and $M$ – the mass of the solvent.

$v = m / M_r$

where $m$ – the mass of sucrose and $M_r$ – the molecular weight of sucrose.

$$v = 222 \, \text{g} / 342 \, \text{g mol}^{-1} = 0.65 \, \text{mol}$$ $$C = 0.65 \, \text{mol} / 1.40 \, \text{kg} = 0.46 \, \text{mol/kg}$$ $$\text{Thus, } \Delta t = K_f \times C = 1.86 \, \text{K kg mol}^{-1} \times 0.46 \, \text{mol/kg} = 0.86 \, \text{K}$$

The freezing point of the solution is $0^{\circ} \text{C} - \Delta t = -0.86^{\circ} \text{C}$.

2. The change of the boiling point is found:

$$\Delta t = K_b \times C$$

where $K_b$ – the embullioscopic constant ($K_b = 0.52 \, \text{K kg mol}^{-1}$ for water) and $C$ – the molality of the solution.

$$\text{Thus, } \Delta t = K_b \times C = 0.52 \, \text{K kg mol}^{-1} \times 0.46 \, \text{mol/kg} = 0.24 \, \text{K}$$

The boiling point is $100^{\circ} \text{C} + \Delta t = 100.24^{\circ} \text{C}$.

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