Calculate the mass in kg of 1.90 L of a 10 mol/L AgClO4
The correct answer and explanation is:
To calculate the mass in kilograms (kg) of 1.90 L of a 10 mol/L solution of AgClO₄ (silver perchlorate), we follow these steps:
Step 1: Determine the number of moles
We are given:
- Volume = 1.90 L
- Molarity (M) = 10 mol/L
Moles=Molarity×Volume=10 mol/L×1.90 L=19.0 mol\text{Moles} = Molarity \times Volume = 10 \, \text{mol/L} \times 1.90 \, \text{L} = 19.0 \, \text{mol}
Step 2: Calculate the molar mass of AgClO₄
Breakdown:
- Ag = 107.87 g/mol
- Cl = 35.45 g/mol
- O₄ = 4 × 16.00 = 64.00 g/mol
Molar mass of AgClO₄=107.87+35.45+64.00=207.32 g/mol\text{Molar mass of AgClO₄} = 107.87 + 35.45 + 64.00 = 207.32 \, \text{g/mol}
Step 3: Calculate the mass in grams
Mass=Moles×Molar mass=19.0 mol×207.32 g/mol=3939.08 g\text{Mass} = \text{Moles} \times \text{Molar mass} = 19.0 \, \text{mol} \times 207.32 \, \text{g/mol} = 3939.08 \, \text{g}
Step 4: Convert grams to kilograms
Mass in kg=3939.08 g1000=3.93908 kg\text{Mass in kg} = \frac{3939.08 \, \text{g}}{1000} = 3.93908 \, \text{kg}
✅ Final Answer: 3.94 kg (rounded to 3 significant figures)
300-Word Explanation
To calculate the mass of a chemical in a solution, we need to understand the concept of molarity and the molar mass of the compound. Molarity (mol/L) expresses how many moles of solute (in this case, silver perchlorate or AgClO₄) are present in one liter of solution. The problem gives a molarity of 10 mol/L and a volume of 1.90 L, meaning we can calculate the total number of moles of AgClO₄ using the formula: moles=molarity×volume\text{moles} = \text{molarity} \times \text{volume}
This gives 19.0 moles of AgClO₄ in 1.90 L of solution.
Next, we need to determine how much this quantity of AgClO₄ weighs. For this, we need its molar mass—which is the mass of one mole of the compound. We calculate this by summing the atomic masses of all the atoms in the molecule:
- Silver (Ag) contributes 107.87 g/mol
- Chlorine (Cl) adds 35.45 g/mol
- Four oxygens (O₄) contribute 64.00 g/mol
Adding these gives a total molar mass of 207.32 g/mol.
We multiply the molar mass by the number of moles to get the mass in grams: 19.0×207.32=3939.08 g19.0 \times 207.32 = 3939.08 \text{ g}
Finally, to convert grams to kilograms, we divide by 1000. This gives approximately 3.94 kg, which is the total mass of AgClO₄ in 1.90 L of this solution.