You are given 350 mL a stock solution of 1.66E-02 M benzoic acid, MW = 122.125 g/mol.
Answer the following questions.
A. What is the number of moles present in the original stock solution?
B. You need to dilute the benzoic acid stock solution to 1.24E-03 M. If you need 455 mL of solution,
what volume of the original stock solution do you need to dilute?
C. What is the number of moles present in the new diluted solution?
D. You mass the solution to be 462.131 g. What is the density of the solution?
E. If the actual density should be 1.05 g/mL. What is your percent error?
It helps us if you keep to the same screen name.
A. moles = M x L
B. mL x M = mL x M
C. see B.
D. density = mass/volume
E. [(your value-1.05)/1.05]*100 = ??
C. moles = M x L.
A. To find the number of moles present in the original stock solution, we can use the formula:
Number of moles = Concentration (in M) x Volume (in L)
Given:
Concentration = 1.66E-02 M (molarity)
Volume = 350 mL = 0.350 L (convert mL to L)
Plugging in the values, we get:
Number of moles = 1.66E-02 M x 0.350 L = 0.00581 moles
Therefore, there are 0.00581 moles present in the original stock solution.
B. To dilute the benzoic acid stock solution to a desired concentration of 1.24E-03 M, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (1.66E-02 M)
V1 = volume of the stock solution to be diluted (unknown)
C2 = desired concentration (1.24E-03 M)
V2 = final volume of the diluted solution (455 mL = 0.455 L)
Plugging in the values, we get:
(1.66E-02 M) x V1 = (1.24E-03 M) x 0.455 L
Solving for V1, we get:
V1 = [(1.24E-03 M) x (0.455 L)] / (1.66E-02 M)
V1 ≈ 0.034 L
Therefore, you need to dilute approximately 0.034 L (34 mL) of the original stock solution.
C. To find the number of moles present in the new diluted solution, we can use the formula:
Number of moles = Concentration (in M) x Volume (in L)
Given:
Concentration = 1.24E-03 M (molarity of the diluted solution)
Volume = 455 mL = 0.455 L (convert mL to L)
Plugging in the values, we get:
Number of moles = 1.24E-03 M x 0.455 L = 5.64E-04 moles
Therefore, there are approximately 5.64E-04 moles present in the new diluted solution.
D. To find the density of the solution, we can use the formula:
Density = Mass / Volume
Given:
Mass = 462.131 g
Volume = 455 mL = 0.455 L (convert mL to L)
Plugging in the values, we get:
Density = 462.131 g / 0.455 L ≈ 1014 g/L
Therefore, the density of the solution is approximately 1014 g/L.
E. To calculate the percent error, we need to compare the actual density with the calculated density using the formula:
Percent Error = [(Actual Value - Calculated Value) / Actual Value] x 100
Given:
Actual Density = 1.05 g/mL = 1050 g/L (convert g/mL to g/L)
Calculated Density = 1014 g/L (calculated in part D)
Plugging in the values, we get:
Percent Error = [(1050 g/L - 1014 g/L) / 1050 g/L] x 100 ≈ 3.43%
Therefore, the percent error is approximately 3.43%.