How many moles of potassium sulfate would you need to prepare 1200 grams of 5.0% sulfate (m/m) solution?
Calculate the volume of solution of 3.6 NH4CI needed to prepare 350.0ml of 1.24 M NH4CI.
How many grams of ammonium chloride are in the solution you created in the above problem.
To answer these questions, we need to use the concepts of moles, molarity, and mass percent.
1. How many moles of potassium sulfate would you need to prepare 1200 grams of 5.0% sulfate (m/m) solution?
To calculate the number of moles of potassium sulfate, we need the mass of the compound and its formula weight.
Step 1: Determine the mass of the sulfate in the solution:
Mass of sulfate = (5.0/100) * 1200 grams = 60 grams
Step 2: Determine the formula weight of potassium sulfate:
Formula weight of potassium sulfate = atomic weight of potassium + atomic weight of sulfur + (4 * atomic weight of oxygen)
= 39.1 + 32.1 + (4 * 16.0) = 174.3 g/mol
Step 3: Calculate the number of moles of potassium sulfate:
Number of moles = Mass / Formula weight
= 60 grams / 174.3 g/mol
≈ 0.344 moles
So, you would need approximately 0.344 moles of potassium sulfate to prepare 1200 grams of 5.0% sulfate solution.
2. Calculate the volume of solution of 3.6 M NH4CI needed to prepare 350.0 ml of 1.24 M NH4CI.
To calculate the volume of solution, we can use the equation: M₁V₁ = M₂V₂, where M₁ is the initial molarity, V₁ is the initial volume, M₂ is the final molarity, and V₂ is the final volume.
Given:
Initial molarity (M₁) = 3.6 M
Initial volume (V₁) = ?
Final molarity (M₂) = 1.24 M
Final volume (V₂) = 350.0 ml
Step 1: Substitute the given values into the equation:
M₁V₁ = M₂V₂
(3.6 M)(V₁) = (1.24 M)(350.0 ml)
Step 2: Solve for V₁:
V₁ = (1.24 M)(350.0 ml) / (3.6 M)
≈ 121.1 ml
So, you would need approximately 121.1 ml of 3.6 M NH4CI solution to prepare 350.0 ml of 1.24 M NH4CI solution.
3. How many grams of ammonium chloride are in the solution you created in the above problem?
To calculate the mass of ammonium chloride, we can use the equation: Mass = Volume * Molarity * Formula weight.
Given:
Volume = 350.0 ml
Molarity = 1.24 M
Formula weight of ammonium chloride = atomic weight of nitrogen + (4 * atomic weight of hydrogen) + atomic weight of chlorine
= 14.0 + (4 * 1.0) + 35.5 = 53.5 g/mol
Step 1: Substitute the given values into the equation:
Mass = (350.0 ml) * (1.24 M) * (53.5 g/mol)
= 23,298 grams
So, there are approximately 23,298 grams of ammonium chloride in the solution created in the above problem.