What is the volume of 6.00 M nitric acid that contains 6.302 g of HNO3 solute (63.02 g/mol)?

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A) 167 mL

B) 60.0 mL

C) 16.7 mL

D) 0.600 mL

E) 1670 mL

2 Answers

  • This is a stoichiometry problem. What needs to happen is that you need to use the numbers available to you to end up with the correct units. So there is 6.302g HNO3 x (1 mol HNO3/63.02g HNO3) x (1 L HNO3/ 6.00 mol HNO3) x (1000 mL HNO3/ 1 L HNO3) this will give you your answer. So that’s, 6.302g of HNO3 divided by the molar mass, so divide by 63.02 g and times by 1 mole HNO3. Then you need to divide by 6.00 mol HNO3 and times by 1 L, as that is the molarity of HNO3 (6.00M) then if you times that number by 1000 you will have your answer in mL

  • Are you sure of the choices?

    Because I got .600 L not mL

    I might be wrong though

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