Asked by Anonymous
What is the required volume in ml to prepare a liter of solution 0.1 mol of nitric acid. molar mass 63 g / mol Density 1.41 g / ml, 75% weight by weight...
Answers
Answered by
DrBob222
The question is somewhat confusing to me. Here is what I think the question is about.
What volume of 75% (w/w) HNO3, density 1.41 g/mL is required to prepare 1 L of 0.1M HNO3.
First determine the M of the 75% HNO3 starting material. That is
1.41 x 1000 x 0.75 x (1/63) = approx 17 M BUT that's just a close estimate. You need to go through with your own calculations.
Then plug this number into the dilution formula to determine the answer.
mL1 x M1 = mL2 x M2
mL1 x 17M = 1000 x 0.1
Solve for mL1. You take that volume of the HNO3 stock solution and add enough water to make 1000 mL.
What volume of 75% (w/w) HNO3, density 1.41 g/mL is required to prepare 1 L of 0.1M HNO3.
First determine the M of the 75% HNO3 starting material. That is
1.41 x 1000 x 0.75 x (1/63) = approx 17 M BUT that's just a close estimate. You need to go through with your own calculations.
Then plug this number into the dilution formula to determine the answer.
mL1 x M1 = mL2 x M2
mL1 x 17M = 1000 x 0.1
Solve for mL1. You take that volume of the HNO3 stock solution and add enough water to make 1000 mL.
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