Chemistry, 01.10.2019 18:30, SundaeSunday

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?

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answered: hosteenimport21

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.

$V_{base} = V_{final} - V_{initial}$

= (15.51 - 1.72) ml

= 13.79 ml

Now, moles of NaOH will be calculated as follows.

No. of moles = $Molarity \times Volume$

= $0.165 M \times 13.79 ml$

= 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 = $\frac{1}{3} \times 2.275 mmol$

= 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.

answered: Guest

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers

answered: Guest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.