An amount of KMnO4 was added to 2500 mL of water. The boiling point of the solution is now 375.25 K. What mass of KMnO4 (in grams) was added to the water? (Kb water = 0.512 °C/m, MW KMnO4 = 158 g/mol, density of water = 1.0 g/mL, boiling point of water = 373.15 K)
The correct answer and explanation is:
To find the mass of KMnO₄ added, we can use the boiling point elevation formula: ΔTb=Kb⋅m\Delta T_b = K_b \cdot m
Where:
- ΔTb\Delta T_b is the boiling point elevation
- KbK_b is the boiling point elevation constant for water (0.512 °C/m)
- mm is the molality of the solution (mol solute / kg solvent)
Step 1: Calculate ΔTb\Delta T_b
ΔTb=375.25 K−373.15 K=2.10 °C\Delta T_b = 375.25\,K – 373.15\,K = 2.10\,°C
Step 2: Calculate molality (m)
m=ΔTbKb=2.100.512≈4.10 mol/kgm = \frac{\Delta T_b}{K_b} = \frac{2.10}{0.512} \approx 4.10 \,\text{mol/kg}
Step 3: Convert 2500 mL of water to kg
Given the density of water is 1.0 g/mL, then: 2500 mL=2500 g=2.500 kg2500\,\text{mL} = 2500\,\text{g} = 2.500\,\text{kg}
Step 4: Calculate moles of KMnO₄
mol=m×kg water=4.10×2.500=10.25 mol\text{mol} = m \times \text{kg water} = 4.10 \times 2.500 = 10.25\,\text{mol}
Step 5: Convert moles to grams
mass=mol×molar mass=10.25×158=1619.5 g\text{mass} = \text{mol} \times \text{molar mass} = 10.25 \times 158 = \boxed{1619.5\,\text{g}}
Final Answer: 1619.5 grams of KMnO4\boxed{1619.5 \text{ grams of KMnO}_4}
Explanation (≈300 words):
The problem involves a classic colligative property: boiling point elevation, which occurs when a solute is added to a solvent. In this case, potassium permanganate (KMnO₄) is added to water, increasing its boiling point.
To solve this, we start by calculating how much the boiling point has changed. The normal boiling point of water is 373.15 K, and the new boiling point is 375.25 K. This gives a boiling point elevation (ΔTb\Delta T_b) of 2.10 °C.
Next, we use the boiling point elevation formula, ΔTb=Kb⋅m\Delta T_b = K_b \cdot m, where KbK_b is a known constant for water (0.512 °C/m). Rearranging the formula allows us to solve for molality, the number of moles of solute per kilogram of solvent. Dividing 2.10 °C by 0.512 °C/m gives us a molality of 4.10 mol/kg.
Since the volume of water used is 2500 mL, and the density of water is 1.0 g/mL, the mass of the water is 2500 g or 2.500 kg. Multiplying the molality by the mass of water in kg gives us the number of moles of KMnO₄: 4.10 mol/kg × 2.500 kg = 10.25 mol.
Finally, we convert moles to grams using the molar mass of KMnO₄ (158 g/mol). Multiplying gives 10.25 mol × 158 g/mol = 1619.5 grams.
This calculation assumes ideal solution behavior and that KMnO₄ does not dissociate significantly in a way that affects colligative properties (which is reasonable for this level of problem).