How many grams of sulfuric acid are present in 35 ml of 2.5 M sulfuric acid?

if it's 2.5M, then you would make a proportion of 2.5 moles/1 L=x moles/.035 L and solve for x, which would be .0875 moles and multiply that by 98 g/1 mole to get 8.575 grams.