how would you make 500.0mL of a 0.200M NaOH solution from a 15.0 M stock NaOH solution??
I assume that you want the volumes rather than the mechanics of carrying it out?
a) How many moles do you need in the 500 ml to have 0.200 M?
b) What volume of 15.0 M would contain the number of moles from a)?
In principle you would transfer this volume to a 500 ml volumetric flask and make the volume upto the line with water. Not so easy to achieve the 3 significant figures implied in the question in practice because of the small volume needed.
To make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M stock NaOH solution, you would follow these steps:
a) Calculate the number of moles needed in the 500 mL to achieve a 0.200 M concentration.
To do this, you need to use the formula:
moles = concentration (M) x volume (L)
The volume is given as 500 mL, so we need to convert it to liters:
500 mL = 500 mL / 1000 mL/L = 0.5 L
Now, plug in the values into the formula:
moles = 0.200 M x 0.5 L = 0.100 moles
Therefore, you need 0.100 moles of NaOH.
b) Calculate the volume of the 15.0 M stock solution that contains the number of moles calculated in a).
To do this, rearrange the formula above to solve for volume:
volume (L) = moles / concentration (M)
Plug in the values:
volume (L) = 0.100 moles / 15.0 M = 0.0067 L
Therefore, you would need 0.0067 L or 6.7 mL of the 15.0 M stock NaOH solution.
In practice, you would transfer this volume to a 500 mL volumetric flask and make the volume up to the line with water. However, keep in mind that achieving the exact concentration of 0.200 M may be challenging due to the small volume needed.