Question

28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cirtic acid are in the samples?

b. How many grams of citric acid are in the sample?












c. What volume of the initial solution would you need make a 250mL solution of 0.0050M citric acid?

Answer 1

#a. Answer:

0.0063 mole

Solution and explanation:

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

= 0.0063 mole

#b. Answer

1.21 g citric acid

Solution

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

= 1.21 g citric acid

#c. Answer

4.167 mL

Solution:

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

= (0.0050 × 0.25)÷ 0.30

= 0.004167 L or

= 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL