9. How much water must be added to 750 mL of 0.332 M
FC) to reduce its concentration to 0.250 M?
Answers
Answered by
3
Answer:
246 ml
Explanation:
Let the water added be x ml
We know that M1V1 = M2V2
HENCE,
750 × 0.332 = 0.250 × ( x + 750) ( since x water is added to 750 ml)
= x + 750 = 750×0.332/0.250
= x + 750 = 996
x = 996 - 750 = 246 ml
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