Solution 1: 20.0 mL x 3.0 M = ?1
Solution 2: 30.0 mL x 5.0 M = ?2
New concentration = ?1 + ?2
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.
4 answers
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)
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.
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.