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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

User Sgiri
by
7.1k points

2 Answers

3 votes

Answer:

0.305

Step-by-step explanation:

0.305

User Arninja
by
7.1k points
5 votes

Answer:

V1= 0.305L

Step-by-step 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=


(0.838 * 0.455)/(1.25 ) = 0.305

User Leetwinski
by
6.7k points