What mass of ethylene glycol (molar mass = 62.0 g mol-1) must be added to 5.50 kg of water to lower the freezing point of water from 0° C to -10.0° C? (K, for water = 1.86 K kg mol-1)?
The Correct Answer and Explanation is :
To solve this problem, we can use the concept of freezing point depression, which is described by the formula: ΔTf=Kf⋅m\Delta T_f = K_f \cdot m
Where:
- ΔTf\Delta T_f is the change in freezing point (in °C),
- KfK_f is the cryoscopic constant (in K kg/mol),
- mm is the molality of the solution (in mol/kg).
Step 1: Calculate the change in freezing point
The freezing point of water is initially 0°C, and it is lowered to -10.0°C. So, the change in freezing point is: ΔTf=0−(−10.0)=10.0 °C\Delta T_f = 0 – (-10.0) = 10.0 \, \text{°C}
Step 2: Use the formula to find the molality
Now, we rearrange the freezing point depression equation to solve for molality: m=ΔTfKfm = \frac{\Delta T_f}{K_f}
Substitute the values: m=10.0 °C1.86 K kg/mol=5.376 mol/kgm = \frac{10.0 \, \text{°C}}{1.86 \, \text{K kg/mol}} = 5.376 \, \text{mol/kg}
Step 3: Calculate the moles of ethylene glycol
The molality of the solution is 5.376 mol of ethylene glycol per kg of water. Given that we have 5.50 kg of water, the moles of ethylene glycol needed is: moles of ethylene glycol=5.376 mol/kg×5.50 kg=29.568 mol\text{moles of ethylene glycol} = 5.376 \, \text{mol/kg} \times 5.50 \, \text{kg} = 29.568 \, \text{mol}
Step 4: Convert moles of ethylene glycol to mass
Now, to find the mass of ethylene glycol, we use the molar mass (62.0 g/mol): mass of ethylene glycol=29.568 mol×62.0 g/mol=1833.2 g\text{mass of ethylene glycol} = 29.568 \, \text{mol} \times 62.0 \, \text{g/mol} = 1833.2 \, \text{g}
Convert grams to kilograms: 1833.2 g=1.833 kg1833.2 \, \text{g} = 1.833 \, \text{kg}
Final Answer:
The mass of ethylene glycol that must be added is 1.83 kg.
Explanation:
The problem requires us to calculate the mass of ethylene glycol needed to lower the freezing point of water. Freezing point depression depends on the molality of the solute, which is the number of moles of solute per kilogram of solvent. Using the known freezing point depression constant for water (1.86 K kg/mol), we calculated the required molality and moles of ethylene glycol. After determining the moles, we multiplied by the molar mass of ethylene glycol to find the mass required.