Chemistry, asked by Brainlystain, 11 months ago

the amount of na2co3 present in 100ml of 0.1M

Answers

Answered by aartigiram22

1

Answer:

83g

Explanation:

n=MV(lit)

n= 0.1×100/1000= 1 mol

weight of 1mol of Na2CO3= 83g

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