How many mL of water must be added to 45.0 mL of a 1.25 M HCL (aq) to lower the concentration to 1.00 M?
Do you just set it up as 45mL/1.25 = X/1.00M and X=36?
That seems too easy to be true. How would you approach this type of question?
2007-08-29 18:58:25 · 2 answers · asked by ecstyle483 1 in Science & Mathematics ➔ Chemistry
You just don't set up these things. You follow basic good sense.
The basic good sense is that the AMOUNT of HCl is unchanged. So you initally have 0.0045 L*1.25 M/L = 5.62x10-3 moles of HCl. To make a 1 M solution, you would add this much HCl to 0.00562 L or 56.2 mL of water. Since you already have 45 mL of solution, you add 11.2 mL of water.
2007-08-29 19:07:53 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
This is a simple case of using the M1V1=M2V2 equality.
45.0 mL*1.25 M = x mL*1.00 M
(45.0 mL*1.25M)/1.00 M = 56.25 mL
To answer the question, find the difference:
56.25 mL - 45.0 mL = 11.25 mL
2007-08-30 02:10:43 · answer #2 · answered by Jeffrey Theta 1 · 0⤊ 0⤋