What volume of 1.60 M SrCL2 is needed to prepare 525 ml of 5.00mM SrCl2?
To find the volume of 1.60 M SrCl2 needed to prepare 525 ml of 5.00 mM SrCl2, we need to use the formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, C1 is 1.60 M, V1 is what we are trying to find, C2 is 5.00 mM (which is equivalent to 0.005 M), and V2 is 525 ml (which is equivalent to 0.525 L).
So, the equation becomes:
(1.60 M)(V1) = (0.005 M)(0.525 L)
To solve for V1, we rearrange the equation:
V1 = (0.005 M)(0.525 L)/(1.60 M)
Calculating this equation will give us the volume:
V1 = 0.00163125 L
To convert this volume to milliliters, we can multiply by 1000:
V1 = 1.63125 ml
Therefore, approximately 1.63125 ml of 1.60 M SrCl2 is needed to prepare 525 ml of 5.00 mM SrCl2.