The molecular weight of Potassium chloride (KCl) is 74.55 g/mol. How many grams would you need for 650 ml of a 35 mM solution?
The Correct Answer and Explanation is:
Correct Answer:
1.7 grams of KCl
Explanation:
To determine how many grams of potassium chloride (KCl) are needed to prepare 650 milliliters (ml) of a 35 millimolar (mM) solution, we follow a step-by-step approach using the molarity formula and molecular weight.
Step 1: Understand the relationship
Molarity (M) is defined as the number of moles of solute per liter of solution.
We are given:
- Volume = 650 ml = 0.650 liters (since 1000 ml = 1 liter)
- Molarity = 35 mM = 0.035 mol/L (since 1000 mM = 1 mol)
- Molecular weight of KCl = 74.55 g/mol
Step 2: Calculate the number of moles of KCl required
Use the formula:
moles = molarity × volume
moles of KCl = 0.035 mol/L × 0.650 L = 0.02275 mol
Step 3: Convert moles to grams
Use the molecular weight to convert moles to grams:
mass (g) = moles × molecular weight
mass = 0.02275 mol × 74.55 g/mol = 1.695 grams
Step 4: Round the answer
Rounding 1.695 grams to 1 decimal place, we get:
1.7 grams
Summary
To prepare 650 ml of a 35 mM potassium chloride (KCl) solution, you would need to weigh out and dissolve 1.7 grams of KCl in enough water to make up 650 ml of solution. This calculation is essential in chemistry labs to ensure solutions have the correct concentration for accurate experimental results.
