How many mL of pure isopropyl alcohol is needed to prepare 250 mL of 70% isopropyl alcohol (C3H7OH). I'm thinking if i should use vol of solute/vol of sol'n but it doesn't fit the formula and im confuse it. please help me
This is just a dilution.
mL x % = mL x %
250mL x 70% = ?mL x 100%
To determine the volume of pure isopropyl alcohol needed to prepare a given volume of a specific concentration, you can use the formula:
V1 × C1 = V2 × C2
Where:
V1 = Volume of pure isopropyl alcohol (unknown)
C1 = Concentration of pure isopropyl alcohol (100% or 1)
V2 = Final volume of the solution (250 mL)
C2 = Desired concentration of isopropyl alcohol (70% or 0.7)
Using the formula, we can rearrange it to solve for V1:
V1 = (V2 × C2) / C1
Substituting the values into the formula:
V1 = (250 mL × 0.7) / 1
V1 = 175 mL
Therefore, you would need 175 mL of pure isopropyl alcohol to prepare 250 mL of a 70% isopropyl alcohol solution.