If you mix 20.0mL of a 3.0M solution with 30mL of a 5.00M solution, determine the concentration in molarity of the new solution.

Solution 1: 20.0 mL x 3.0 M = ?1

Solution 2: 30.0 mL x 5.0 M = ?2

New concentration = ?1 + ?2

Isn't this less than the original concentrations?

First, convert mL to L:

20mL=0.0200L and 30.0mL=0.0300L

Solve for moles:

M=moles/Volume (L) so, moles=M*Volume(L)

New Molarity=Total moles/Total Volume

M=[(0.0200L)*(3.0 M)+(0.0300L)*(2.0M)]/(0.0200L+0.0300L)

I'll get in on the fray here. Irving has it wrong. Devron made a typo. I've copied Devron's work and replaced the typo part with a bold 5.0

M=[(0.0200L)*(3.0 M)+(0.0300L)*(5.0M)]/(0.0200L+0.0300L)

The answer comes out to be about 4.2 which is the weighted average between the 3.0 and 5.0 M which makes sense. You know it must be somewhere between 3.0 M and 5.0 M.

To determine the concentration in molarity of the new solution, you will need to calculate the total amount of moles of solute and the total volume of the solution. Then, divide the moles by the volume to get the concentration in molarity.

Step 1: Calculate the moles of solute for each solution:
For the first solution:
Moles of solute = concentration in M (mol/L) × volume in L
Moles of solute = 3.0 mol/L × 0.020 L = 0.06 mol

For the second solution:
Moles of solute = concentration in M (mol/L) × volume in L
Moles of solute = 5.00 mol/L × 0.030 L = 0.15 mol

Step 2: Calculate the total moles of solute by adding the moles for each solution:
Total moles of solute = moles of first solution + moles of second solution
Total moles of solute = 0.06 mol + 0.15 mol = 0.21 mol

Step 3: Calculate the total volume of the solution by adding the volumes of each solution:
Total volume of solution = volume of first solution + volume of second solution
Total volume of solution = 0.020 L + 0.030 L = 0.050 L

Step 4: Calculate the concentration in molarity of the new solution:
Concentration (Molarity) = Total moles of solute / Total volume of solution
Concentration (Molarity) = 0.21 mol / 0.050 L = 4.20 M

Therefore, the concentration in molarity of the new solution is 4.20 M.