What volume of WATER MUST BE ADDED to 10.0 mL of 6.0 M NaOH to make a solution that is 0.30 M in NaOH? Assume that the volumes are additive.
a. 10 mL
b. 190 mL
c. 200 mL
d. 210 mL
e. 500 mL
190
Well, let's do some math and add a touch of comedy to it, shall we?
To find the volume of water we need to add, we can use the equation:
(initial volume of NaOH) × (initial molarity of NaOH) = (final volume) × (final molarity of NaOH)
So, let's plug in the values:
(10.0 mL) × (6.0 M) = (final volume) × (0.30 M)
Now, let's solve for the final volume:
(final volume) = (10.0 mL × 6.0 M) / (0.30 M)
Calculating this, we get:
(final volume) = 200 mL
Voila! The answer is c. 200 mL. So, it looks like you'll need to add 200 mL of water to your NaOH solution.
And remember, if you're feeling a little parched, don't drink the NaOH solution. Stick to regular water instead!
To determine the volume of water that must be added to the 10.0 mL of 6.0 M NaOH solution to make a solution that is 0.30 M in NaOH, we can use the concept of dilution. The formula for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
In this case, we can rearrange the formula to solve for V2:
V2 = (C1V1) / C2
Plugging in the values:
C1 = 6.0 M
V1 = 10.0 mL
C2 = 0.30 M
V2 = (6.0 M * 10.0 mL) / 0.30 M
V2 = 200 mL
Therefore, the volume of water that must be added is 200 mL. The correct answer is option c.
To find the volume of water that must be added, we need to use the concept of dilution.
The formula for dilution is:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Given:
C1 = 6.0 M
V1 = 10.0 mL
C2 = 0.30 M
We need to find V2, the final volume (which includes both the volume of NaOH and the volume of water added).
Using the formula for dilution, we can rearrange it to solve for V2:
V2 = (C1 * V1) / C2
V2 = (6.0 M * 10.0 mL) / 0.30 M
V2 = 60 mL / 0.30 M
V2 = 200 mL
Therefore, the volume of water that must be added is 200 mL.
The correct answer is c. 200 mL.
you are diluting it 20 times? Add 1 part original, 19 parts water.
Now if you want to do it the hard way:
molesacid=10ml*6M=V*.3
Volumetotal=200ml
so water added =200-10ml