Question

How many moles of nitric acid are present in 35.0 ml of a 2.20 M solution?

Answer 1

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Step-by-step explanation:

Molarity of the solution = 2.20 M

`Molarity=(number\:of\:moles)/(Volume\:of\:Solution\:in\:L)\\\\Number\:of\:moles=Molarity*(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35*10^(-3)L\\\\Number\:of\:moles=2.20*35*10^(-3)=77*10^(-3)\:moles\:of\:HNO_(3)`

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution