2. 2mL of a 10M solution of Compound A is taken and 48mL added. 1mL of this solution is then taken and dried to dryness before being re-dissolved in 0.1mL. What is the concentration of the 0.1mL solution?
First dilution is 10M x (2/50) = 0.4M
You take 1 mL of that so you take moles = 0.4M x 0.001L = 4E-4 moles.
Then you dissolve that is 0.001 L so M = moles/L which is 4E-4/0.001 = ?M
To find the concentration of the 0.1 mL solution, we need to calculate the amount of Compound A transferred from the initial solution to the final solution.
Let's break down the problem step by step:
Step 1: Initial solution preparation
- We start with 2 mL of a 10 M solution of Compound A.
- This means we have 10 moles of Compound A in 1 liter of solution.
- Therefore, in 2 mL of solution, we have (10 mol/L) x (2 mL/1000 mL) = 0.02 moles of Compound A.
Step 2: Dilution
- We add 48 mL of a solvent (probably water) to the initial 2 mL solution.
- This increases the total volume to 2 mL + 48 mL = 50 mL.
- However, the moles of Compound A remain the same at 0.02 moles.
Step 3: Calculation of concentration
- We take 1 mL of the diluted solution and dry it to dryness.
- When the solution is dried to dryness, all the solvent evaporates, and only the solute remains.
- This means that the 1 mL of solution now contains the same amount of Compound A, 0.02 moles.
- We then re-dissolve this dry residue in 0.1 mL of solvent, which is a much smaller volume.
- Since the moles of Compound A remain constant, dividing the amount by the volume gives us the concentration.
- Thus, the concentration of the 0.1 mL solution is (0.02 moles)/(0.1 mL) = 0.2 M.
Therefore, the concentration of the 0.1 mL solution is 0.2 M.