Question

What molarity should the stock solution be if you ant to dilute 25.0 mL to 2.00 L and have the final concentration be 0.103M? a. 4.12M, b. 8.24 M c. 0.243 M, d. 0.206M?

Answer 1

To find the molarity of the stock solution, the dilution equation C1V1 = C2V2 is used. After converting and plugging in the values, the molarity of the stock solution required is determined to be 8.24 M. (Option b)

Step-by-step explanation:

The student is asking what molarity the stock solution should be if they want to dilute 25.0 mL of it to 2.00 L and achieve a final concentration of 0.103 M. The calculation to find the concentration of the stock solution (C1) uses the dilution equation C1V1 = C2V2, where V1 is the volume of the stock solution (25.0 mL) and C2 is the final concentration (0.103 M), and V2 is the final volume (2.00 L).

Converting 2.00 L to mL gives us 2000 mL to keep the units consistent. Plugging in the known values gives us:

1(25.0 mL) = (0.103 M)(2000 mL)

Now we can solve for C1:

C1 = (0.103 M)(2000 mL) / 25.0 mL = 8.24 M

Therefore, the correct answer is (b) 8.24 M.

Answer 2

A common dilution relationship is (M1V1)=(M2V2)
(0.103M x 2.00L) = (M2 x 0.025L)
M2= (0.103M x 2.00L)/0.025L
M2= 8.24M