Zacznij za darmoZacznij

Zadanie 2c

4 pktobliczenioweśrednie (szac.)Paper 3, maj 2023, TZ2

Wzorowane na: Chemistry SL Paper 3, May 2023, TZ2, pytanie 2(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

Zapisz zadanie

Darmowe konto pozwoli wrócić do niego później.

Informacja do zadań 2a-2d.

A student investigated how much spinach would supply the recommended daily intake of iron. The student

- weighed 63.2 g of spinach leaves and blended them with 600 cm3 of water

- boiled the mixture, filtered it and allowed the filtrate to cool

- pipetted 10.0 cm3 of the filtrate into 20.0 cm3 of 1.00 mol dm−3 sulfuric acid in a flask

- titrated the mixture with 0.00100 mol dm−3 potassium manganate(VII) solution.

The reaction taking place is:

5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)

Polecenie

The end point was reached after 3.0 ± 0.1 cm3 of titrant had been run in.

(i) Calculate the percentage uncertainty in this titre.

(ii) Suggest one change to the procedure that would lower the percentage uncertainty of a single titration, apart from fitting a more precise burette.

(iii) The titrated solution contained 8.38 × 10−4 g of iron. Determine the percentage of iron by mass in the spinach leaves, giving your answer to three significant figures.

Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja

Zanim porównasz z kluczem

Jak poszło Ci z tym zadaniem?

Wybierz ocenę: zapisze wynik i od razu oznaczy zadanie jako zrobione.

ŁatweŚrednieTrudne

Zapis wymaga darmowego konta. Załóż konto

Odpowiedź z klucza

(i) «0.13.0×100=\dfrac{0.1}{3.0} \times 100 =» 3.3 «%» ✔

(ii) any one of:

using a more dilute potassium manganate(VII) solution «so that the titre is larger» ✔

OR

using a larger aliquot/volume of filtrate ✔

OR

using a larger mass of spinach/less water for the extraction «so that the filtrate is more concentrated» ✔

(iii) ALTERNATIVE 1

mass of iron in the 63.2 g of spinach «=8.38×10−4×60010.0== 8.38 \times 10^{-4} \times \dfrac{600}{10.0} =» 0.0503 «g» ✔

percentage by mass «=0.050363.2×100== \dfrac{0.0503}{63.2} \times 100 =» 0.0796 «%» ✔

ALTERNATIVE 2

mass of spinach in the titration flask «=63.2×10.0600== 63.2 \times \dfrac{10.0}{600} =» 1.053 «g» ✔

percentage by mass «=8.38×10−41.053×100== \dfrac{8.38 \times 10^{-4}}{1.053} \times 100 =» 0.0796 «%» ✔

Schemat punktowania

(i) [1]: accept 3 %. (ii) [1]: any one modification. Do not accept "use a more precise pipette" or "use a better balance", which do not change the percentage uncertainty of the titre. (iii) [2]: award [2] for a correct final answer (0.0796 %) with working. M2 must be to 3 significant figures. Award [1 max] for 0.00133 %, which ignores the dilution of the extract.

Umiejętności w zadaniu