You have a stock solution of 15.0 M NH3 . How many milliliters of this solution should you dilute to make 1200ml of 0.280 M NH3 ?
find the moles of NH3 you need:
moles=.280*volume=.280*1.2=xxxx liters
Now, find the volume of stock solution needed to deliver those moles..
volume=xxxx/15
so take 1000*xxxx/15 ml of stock solution, and the remainder water.
stir, properly label, and you are done.
To solve this problem, you can use the equation for dilution:
M1V1 = M2V2
Where:
M1 = initial molarity (15.0 M)
V1 = initial volume (unknown)
M2 = final molarity (0.280 M)
V2 = final volume (1200 mL or 1.2 L)
First, rearrange the equation to solve for V1:
V1 = (M2 * V2) / M1
Now plug in the given values:
V1 = (0.280 M * 1.2 L) / 15.0 M
Calculate the result:
V1 = 0.0224 L
Remember, since 1 L = 1000 mL, convert liters to milliliters:
V1 = 0.0224 L * 1000 mL/L
V1 = 22.4 mL
Therefore, you need to dilute 22.4 mL of the 15.0 M NH3 solution to make 1200 mL of 0.280 M NH3 solution.