Calculate the concentration, in molarity, of a solution prepared by adding 9 mL of water to 1 mL of 0.1 M HCl solution.
If 2.0 mL of 0.010 M NaOH is mixed with enough water to make the total volume 8 mL, what is the molarity of the resulting solution?
ASSUMING that the 9 mL + the 1 mL adds to 10 mL of solution (which may not be correct), then you have 0.1M x 1/10 = ?
To calculate the concentration, in molarity, of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters).
1) For the first question, the initial volume of the solution is 1 mL (0.001 L) and the initial concentration of HCl is 0.1 M.
To find the moles of HCl, you can use the formula:
moles = concentration x volume
moles = 0.1 M x 0.001 L
moles = 0.0001 mol
Since we dilute the solution by adding 9 mL of water, the final volume becomes 10 mL (0.01 L).
Now, to find the concentration of the resulting solution, divide the moles of HCl by the final volume:
concentration = moles / volume
concentration = 0.0001 mol / 0.01 L
concentration = 0.01 M
Therefore, the concentration of the resulting solution is 0.01 M.
2) For the second question, the initial volume of the solution is 2 mL (0.002 L) and the initial concentration of NaOH is 0.010 M.
Using the same formula as before:
moles = concentration x volume
moles = 0.010 M x 0.002 L
moles = 0.00002 mol
By adding enough water to make the total volume 8 mL, the final volume becomes 8 mL (0.008 L).
To find the concentration of the resulting solution, divide the moles of NaOH by the final volume:
concentration = moles / volume
concentration = 0.00002 mol / 0.008 L
concentration = 0.0025 M
Therefore, the molarity of the resulting solution is 0.0025 M.