Cedrick & astrid titrated a 20.00 ml aliquot of grapefruit juice with a 0.165 m naoh solution to the end point. the initial buret reading was 1.72 ml and the final buret reading was 15.51 ml. they calculated that there was 0.1457 g of citric acid present in the juice sample. what is the amount mg of citric acid present per ml of juice?

Explanation:

Citric acid upon dissociation given three hydrogen ion that is, citric acid is a triprotic acid.

It is given that volume of juice is 20 ml and concentration of NaOH is 0.165 M.

    Initial volume = 1.72 ml,       Final volume = 15.51 ml

Hence, volume of the base will be calculated as follows.

                              = (15.51 - 1.72) ml

                              = 13.79 ml

Now, moles of NaOH will be calculated as follows.

                 No. of moles =

                                       =

                                       = 2.275 mmol

As,   1 mol of  citric acid = 3 mol of NaOH

Hence, moles of citric acid will be calculated as follows.

                 Mole of citric acid =

                                                = 0.758 mmol

Since, molar mass of citric acid is 192.124 g/mol. Hence, mass of citric acid will be calculated as follows.

                      Mass = No. of moles × Molar mass

                                = 0.758 mmol × 192.124 g/mol

                                = 145.72 mg

Thus, we can conclude that there is 145.72 mg of citric acid present per mL of juice.

explanation: no but i do not usually comment

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.