HCL acid is available as a stock solution with a concentration of 10 mol/L. If you need (9.0*10^-1) L of (2.50*10^0) mol/L hydrochloric acid, approximately how much distilled water will you need to make the dilution?
You do this just like the other one. Post it if you have a specific question about what you don't understand.
Yes but the answer was incorrect
can you please show me the proper way to do it?
Thanks :D
To calculate the amount of distilled water needed to make the dilution, we can use the equation for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Given:
C1 = 10 mol/L (concentration of the stock solution)
V1 = unknown (volume of the stock solution)
C2 = 2.50 * 10^0 mol/L (final concentration required)
V2 = 9.0 * 10^-1 L (final volume required)
Rearranging the equation, we can solve for V1:
V1 = (C2 * V2) / C1
Substituting the given values:
V1 = (2.50 * 10^0 mol/L * 9.0 * 10^-1 L) / 10 mol/L
Simplifying, we have:
V1 = 2.25 * 10^-1 L
Therefore, you would need 2.25 * 10^-1 L (or 0.225 L) of the stock solution.
To determine the amount of distilled water required to make the dilution, subtract the volume of the stock solution from the final volume desired:
Volume of distilled water = V2 - V1
Volume of distilled water = 9.0 * 10^-1 L - 2.25 * 10^-1 L
Volume of distilled water = 6.75 * 10^-1 L
So, approximately 6.75 * 10^-1 L (or 0.675 L) of distilled water is needed to make the dilution.