A chemist has a solution of NaOH 4 mol/L and wishes to prepare 500 mL of a
Solution at 3 mol/L. What volume should it use from the 4 mol/L solution?
500 mL of a 3 M solution contains 1.5 moles of solute
volume = 1.5 moles / (4 mol/L)
R_Scott is right in my opinion
You may use the dilution formula, also, which is
mL1 x M1 = mL2 x M2
mL1 x 4M = 500 x 3M
Then mL of the 4M = 500 x 3/4 = ?
To determine the volume of the 4 mol/L solution needed, we can use the concept of dilution. The formula for dilution is:
C1 * V1 = C2 * V2
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 know:
C1 = 4 mol/L (concentration of the 4 mol/L solution)
V1 = unknown (volume of the 4 mol/L solution to be used)
C2 = 3 mol/L (desired concentration of the final solution)
V2 = 500 mL (desired final volume of the solution)
Substituting the values into the dilution formula, we get:
4 mol/L * V1 = 3 mol/L * 500 mL
Simplifying the equation:
V1 = (3 mol/L * 500 mL) / 4 mol/L
V1 = 1500 mL•mol/L / 4 mol/L
V1 = 375 mL
Therefore, the chemist should use 375 mL of the 4 mol/L NaOH solution to prepare 500 mL of a 3 mol/L NaOH solution.