How many grams of potassium sulfate are needed to prepare 150 ml of a 6.50 m/v% potassium sulfate solutions
To determine the number of grams of potassium sulfate needed to prepare a solution, you need to know the volume and the concentration of the solution.
Given:
Volume of solution (V) = 150 ml
Concentration of the solution (m/v%) = 6.50 m/v%
Step 1: Convert the volume from milliliters to liters.
1 ml = 0.001 L
150 ml = 150 x 0.001 L
150 ml = 0.15 L
Step 2: Recognize that the concentration in m/v% means grams of solute per 100 ml of solution. Therefore, a 6.50 m/v% potassium sulfate solution contains 6.50 grams of potassium sulfate per 100 ml of solution.
Step 3: Calculate the number of grams of potassium sulfate needed for 0.15 L of the solution.
6.50 grams/100 ml = x grams/0.15 L
To solve the proportion:
x = (6.50 grams/100 ml) * 0.15 L
Step 4: Calculate the value of x.
x = (6.50 grams/100 ml) * 0.15 L
x = (6.50 grams * 0.15 L) / 100 ml
x = 0.975 grams
Therefore, you will need 0.975 grams of potassium sulfate to prepare 150 ml of a 6.50 m/v% potassium sulfate solution.