What volume (in L) of a 1.25 M
potassium fluoride (KF) solution
would be needed to make 455 mL of
a 0.838 M solution by dilution?
[?]LKF

Respuesta :

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

[tex] \frac{0.838 \times 0.455}{1.25 } = 0.305 [/tex]

Answer:

0.305

Explanation:

0.305

ACCESS MORE
EDU ACCESS
Universidad de Mexico