If 125 mL of a 6.00 M KNO3 solution is diluted with 100.0 mL of water, the molarity of the resulting solution would be
3.33 M
ASSUMING volumes are additive, which isn't exactly so, it will be
6.00M x (125/225) = ?
or use the dilution formula
mL1 x M1 = mL2 x M2
100*6.00 = 225*M2.
Solve for M2.
To find the molarity of the resulting solution, we need to use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of the solution before dilution (6.00 M)
V1 = initial volume of the solution before dilution (125 mL)
M2 = final molarity of the resulting solution
V2 = final volume of the resulting solution
In this problem, the initial volume (V1) is 125 mL and the final volume (V2) is the sum of the initial volume and the volume of water added (100.0 mL). Therefore, V2 = 125 mL + 100.0 mL = 225.0 mL.
Now we can substitute the given values into the formula:
(6.00 M)(125 mL) = M2(225.0 mL)
Simplifying the equation further:
750 mL·M = 225.0 mL·M2
Now we can solve for M2 by dividing both sides of the equation by 225.0 mL:
M2 = (6.00 M)(125 mL) / 225.0 mL
M2 = 3.33 M
Therefore, the molarity of the resulting solution would be 3.33 M.