a 30.0 ml solution of potassium nitrate was diluted to 125 ml and 25 ml of this solution were then diluted to 150 ml. the concentration of the final solution is 0.00307 M. Calculate the concentration of the original solution.
xM*(30/125)*(25/150) = 0.00307
i do not know if this is right but this is how i would do it.
M1V 1=M2V2.
(0.00307 M) X (150ml)= (M2) X (30 ml)
and then you solve to get m2
Sarah's method will work if you do it TWICE; once for each dilution.
0.00307*150 = 25*c
c = ??
??*125 = 30c
c = xx.
You should arrive at the same answer either way.
To calculate the concentration of the original solution, we need to use the concept of dilution. Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent. The key equation for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (unknown)
V1 = initial volume of the solution (30.0 ml)
C2 = final concentration of the solution (0.00307 M)
V2 = final volume of the solution (150 ml)
First, let's find the concentration of the 25 ml solution that was diluted to 150 ml:
C1 x 25 ml = 0.00307 M x 150 ml
Rearranging the equation, we get:
C1 = (0.00307 M x 150 ml) / 25 ml
C1 = 0.01842 M
So, the concentration of the 25 ml solution before dilution is 0.01842 M.
Now, let's find the concentration of the original solution. We can use the same dilution equation:
C1 x 30.0 ml = 0.01842 M x 125 ml
Rearranging the equation, we get:
C1 = (0.01842 M x 125 ml) / 30.0 ml
C1 = 0.076675 M
Therefore, the concentration of the original solution is approximately 0.0767 M.