what volume of 2.0 M NaCl is required to prepare 150 mL of 0.10 M NaCl?
To determine the volume of 2.0 M NaCl required to prepare 150 mL of 0.10 M NaCl solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = Initial concentration
V1 = Initial volume
C2 = Final concentration
V2 = Final volume
Now let's plug in the given values into the formula and solve for V1:
C1 = 2.0 M
V1 = ?
C2 = 0.10 M
V2 = 150 mL
2.0 M * V1 = 0.10 M * 150 mL
Now, let's solve for V1:
V1 = (0.10 M * 150 mL) / 2.0 M
V1 = (0.10 * 150) / 2.0
V1 = 15 mL
Therefore, you will need 15 mL of 2.0 M NaCl to prepare 150 mL of 0.10 M NaCl solution.
To find the volume of 2.0 M NaCl required to prepare 150 mL of 0.10 M NaCl, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, C1 is 2.0 M, V1 is unknown, C2 is 0.10 M, and V2 is 150 mL.
Substituting the values into the dilution formula, we get:
(2.0 M)(V1) = (0.10 M)(150 mL)
Now, solve for V1:
V1 = (0.10 M)(150 mL) / 2.0 M
V1 = (0.10)(0.150 L) / 2.0
V1 = 0.015 L = 15 mL
Therefore, 15 mL of 2.0 M NaCl is required to prepare 150 mL of 0.10 M NaCl.
How many mols NaCl do you want? That's M x L = ?
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.