Question

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

Answer 1

Answer:

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

Explanation:

Molarity of the solution = 2.20 M

$Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}$

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