0ml of a 6.0 m hno3 stock solution is diluted using water to 100 ml. how many moles of hno3 are present in the dilute solution? 25.0 ml of a 6.0 m stock solution is diluted using water to 100 ml. how many moles of are present in the dilute solution? 6.0 moles 0.6 moles 0.15 moles 1.5 moles

When you are dealing with mixtures that are non-reactive such as HNO3 or nitric acid and water, you do a process of dilution. All you did was decrease the concentration of the HNO3 solution, but it still contains the same amount of moles for the solute. In the concept of law of conservation of mass, the amount of mass or moles (if non-reactive) are additive, no more no less. The final moles would then be the original moles of solute and the added water. 

You can solve this by multiplying the concentration in molarity with the volume. Molarity is moles of solute per liter solution. Since the volume is 100 mL or 0.1 L, then

6 moles/L * 0.1 L = 0.6 moles nitric acid

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Answer is to avoid experimental error