how many milliliters of 5M NaCl are required to prepare 1500 ml of .002M solution?
moles of solution you want? M x L = 0.002 x 1.5 L = ??
That will be the moles you need. Where do you get that? From 5 M NaCl. moles = M x L.
You know moles and M, calculate Liters and change to mL.
You can also use the formula
mL x M = mL x M
mL x 5 = 1500 x 0.002
Well, let's break it down. If you want a 0.002M solution and you have a 5M solution, you're going to need to dilute the heck out of that NaCl. It's like trying to turn a roaring lion into a timid little kitten.
To calculate the amount of concentrated solution needed, you can use the equation: M₁V₁ = M₂V₂.
In this case, your concentrated solution has a concentration of 5M, and you want to end up with a 0.002M solution with a volume of 1500 ml.
So, plugging the values into the equation and solving for V₁:
(5M)(V₁) = (0.002M)(1500 ml)
Now, let's not forget to divide by 5M to isolate V₁:
V₁ = (0.002M)(1500 ml) / 5M
Crunching the numbers, we find that V₁ is equal to 0.6 ml.
So, you would need approximately 0.6 milliliters of your 5M NaCl solution to prepare 1500 ml of a 0.002M solution. It's a small amount, but hey, sometimes less is more!
To calculate the volume of 5M NaCl needed to prepare a 0.002M solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = desired concentration of the diluted solution
V2 = desired volume of the diluted solution
Let's substitute the given values into the formula:
C1 = 5M
V1 = unknown
C2 = 0.002M
V2 = 1500ml
5M * V1 = 0.002M * 1500ml
Solving for V1:
V1 = (0.002M * 1500ml) / 5M
V1 = (0.002 * 1500) / 5
V1 = 0.6 ml
Therefore, you would need to measure 0.6 milliliters of 5M NaCl to prepare 1500 ml of a 0.002M solution.
To determine the number of milliliters of 5M NaCl required to prepare a 1500 ml 0.002M solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (5M NaCl)
V1 = volume of the stock solution needed
C2 = desired concentration of the final solution (0.002M NaCl)
V2 = total volume of the final solution (1500 ml)
Rearranging the formula to solve for V1, we have:
V1 = (C2 × V2) / C1
Substituting the values into the formula:
V1 = (0.002M × 1500ml) / 5M
V1 = (0.003) / 5
V1 = 0.0003 L or 0.3 ml
Therefore, to prepare a 1500 ml 0.002M solution of NaCl, you would need to measure 0.3 ml of the 5M NaCl stock solution.