How much of each starting material would you use to prepare 2.00 L of each of the following solutions?
0.240 M NaOH from 1.00 M NaOH stock solution
what do i do with the .100 M and .240 M??
For dilution problems (preparing a weaker solution from a stronger stock solution) use:
V1M1 = V2M2
(1.00M)(V1) = (0.240M)(2.00L)
Solving for V1, V1 = (0.240M)(2.00L) / (1.00M) = ?
The second part, I don't quite follow.
Well, if we want to prepare 2.00 L of a 0.240 M NaOH solution using a 1.00 M NaOH stock solution, we need to do some calculations.
To convert the 0.240 M solution to a 1.00 M solution, we can use the equation:
C1V1 = C2V2
Where C1 is the concentration of the stock solution (1.00 M), V1 is the volume of the stock solution we need to use (unknown), C2 is the desired concentration (0.240 M), and V2 is the desired final volume (2.00 L).
Plugging in the values, we get:
(1.00 M)(V1) = (0.240 M)(2.00 L)
Solving for V1, we get:
V1 = (0.240 M)(2.00 L) / (1.00 M)
V1 = 0.480 L
So, to prepare 2.00 L of a 0.240 M NaOH solution from a 1.00 M NaOH stock solution, you would need to measure out 0.480 L of the stock solution and then add water to reach a final volume of 2.00 L.
As for what to do with the 0.100 M solution, well... you could always use it to water your plants and give them a little sodium hydroxide boost! Just kidding, please don't do that. The 0.100 M solution could be used for other experiments or disposed of properly according to your laboratory's guidelines.
To prepare 2.00 L of a 0.240 M NaOH solution from a 1.00 M NaOH stock solution, you can use the formula:
M1V1 = M2V2
Where:
M1 = initial concentration of the NaOH stock solution
V1 = volume of the NaOH stock solution you need to measure
M2 = final concentration of the desired NaOH solution (0.240 M)
V2 = final volume of the desired NaOH solution (2.00 L)
First, substitute the known values into the equation:
(1.00 M)(V1) = (0.240 M)(2.00 L)
Next, solve for V1 (the volume of the stock solution you need to measure):
(1.00 M)(V1) = (0.240 M)(2.00 L)
V1 = (0.240 M)(2.00 L) / 1.00 M
V1 = 0.480 L
Therefore, you need to measure 0.480 L (or 480 mL) of the 1.00 M NaOH stock solution and then dilute it with water to a final volume of 2.00 L to obtain a 0.240 M NaOH solution.
To prepare 2.00 L of a 0.240 M NaOH solution from a 1.00 M NaOH stock solution, you need to follow a dilution formula. The formula 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.
In this case, the initial concentration (C1) is 1.00 M, and the final concentration (C2) is 0.240 M. The final volume (V2) is 2.00 L.
To find the initial volume (V1) of the stock solution needed, plug in the values into the formula:
C1V1 = C2V2
(1.00 M) V1 = (0.240 M) (2.00 L)
V1 = (0.240 M) (2.00 L) / (1.00 M)
V1 = 0.480 L
So, to prepare 2.00 L of a 0.240 M NaOH solution from a 1.00 M NaOH stock solution, you would need to measure 0.480 L of the 1.00 M NaOH stock solution and then dilute it to a final volume of 2.00 L with solvent (usually water).
As for the 0.100 M solution mentioned, it is not directly relevant to preparing the 0.240 M solution from the 1.00 M stock solution. If you have any other specific questions regarding the 0.100 M solution, feel free to ask.