Chemistry, asked by rathnakarpandith, 5 hours ago

what is the molarity of solution prepared 200ml of 0.4m of sulphuric acid solution with 1000ml of H2O​

Answers

Answered by ramamsc2007
0

Explanation:

M1V1 = M2V2

M1 = Molarity of sulphuric acid = 0.4 M

V1 of H2SO4 = 200 mL

M2 = Molarity of solution to be prepared= ?

V2 of new solution to be prepared= 1000 mL

M2 = M1V1 / V2 = 0.4 x 200 / 1000 = 0.4 / 5 = 0.08 M

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