How many moles of C12H22O11C12H22O11are needed to prepare 2.50 L of 0.300 M solution?

asked Oct 15, 2021 52.5k views

3 votes

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4 votes

Answer:

0.75 mol

Step-by-step explanation:

concentration=mole/volume

c×v=n

volume=2.50l=2.50dm³

0.300×2.50=0.75mol

Nirvana answered Oct 20, 2021

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