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3 years ago

How many milliliter of a solution of 4.00KI are needed to prepare 250.0mL of 0.760 KI

1 answer:

3 0

Answer:

47.5 mL

Solving:

M1 = 4.00 M

V1 = ?

M2 = 0.760 M

V2 = 0.250 L

---

M1 * V1 = M2 * V2

V1 = ( M2 * V2 ) / M1

V1 = ( 0.760 * 0.250 ) / 4.00

V1 = ( 0.190 ) / 4.00

V1 = 0.0475 L

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