Asked by william
The volume of a H2SO4 solution to 36.75% by weight of density 1, 8g/cm ^ 3, necessary to prepare 500 ml of a solution 3N of sulphuric acid is approximately:
Answers
Answered by
DrBob222
What is the molarity of the initial H2SO4? That's
1.8 g/mL x 1000 mL x 0.3675 x (1/98) = y
N = 2*M
Then mL1 x N1 = mL2 x N2
mL1 x yN = 500 x 3N
Substitute and solve for mL1.
1.8 g/mL x 1000 mL x 0.3675 x (1/98) = y
N = 2*M
Then mL1 x N1 = mL2 x N2
mL1 x yN = 500 x 3N
Substitute and solve for mL1.
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