Question

9. How much water must be added to 750 mL of 0.332 M
FC) to reduce its concentration to 0.250 M?​

Answer 1

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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