How many moles of NaOH are present in 15.5ML of 0.290 M NaOH
No, Therefore, there are 0.004495 moles of NaOH present in 15.5 mL of 0.290 M NaOH.
4.495 moles of NaOH are present in 15.5 mL of 0.290 M NaOH.
To calculate this, you can use the formula:
moles of solute = concentration (M) x volume (L)
First, convert the volume of NaOH from milliliters to liters:
15.5 mL = 0.0155 L
Then, plug in the given concentration and volume values into the formula:
moles of NaOH = 0.290 M x 0.0155 L = 0.004495 moles
Therefore, there are 4.495 moles of NaOH present in 15.5 mL of 0.290 M NaOH.