Zadanie 1e
Wzorowane na: Chemistry SL Paper 2, November 2022, pytanie 1(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 1a-1f.
Ammonium chloride, NH4Cl, is a white crystalline solid used as a nitrogen fertilizer for rice crops and as an electrolyte in dry cells.
Polecenie
A student investigates the dissolution of ammonium chloride in water as the basis of an instant cold pack, in which solid ammonium chloride is kept apart from water by a thin membrane. The enthalpy change when one mole of solid ammonium chloride dissolves in water is +14.8 kJ mol−1.
(i) The mass of the contents of the cold pack is 51.64 g and its initial temperature is 24.8 °C. Once the contents are mixed, the temperature falls to 11.9 °C. Calculate the energy, in J, absorbed by the dissolution of ammonium chloride in water within the cold pack. Assume the specific heat capacity of the solution is 4.18 J g−1 K−1. Use section 1 of the data booklet.
(ii) Determine the mass of ammonium chloride in the cold pack using your answer to (i), the enthalpy change given above and section 7 of the data booklet.
If you did not obtain an answer in (i), use 3.25 × 103 J, although this is not the correct answer.
(iii) The balance used to find the mass of the contents of the cold pack has an absolute uncertainty of ±0.01 g, and each temperature reading has an absolute uncertainty of ±0.2 °C. Calculate the absolute uncertainty in the mass of ammonium chloride in the cold pack, using your answer to (ii).
If you did not obtain an answer in (ii), use 7.80 g, although this is not the correct answer.
(iv) The cold pack actually contains 11.00 g of ammonium chloride. Calculate the percentage error in the experimentally determined mass of ammonium chloride obtained in (ii).
If you did not obtain an answer in (ii), use 7.80 g, although this is not the correct answer.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «ΔT = 24.8 − 11.9 = 12.9 K»
«q = mcΔT = 51.64 g × 4.18 J g−1 K−1 × 12.9 K =» 2.78 × 103 «J» ✔
(ii) « =» 0.188 «mol» ✔
«m = 0.188 mol × 53.50 g mol−1 =» 10.1 «g» ✔
(iii) «fractional / % uncertainty in » = 0.031 / 3.1«%» ✔
«fractional / % uncertainty in » = 0.0002 / 0.02«%»
OR
fractional / % uncertainty in m is much smaller than the uncertainty in ΔT ✔
«3.1% × 10.1 g =» 0.31 «g» ✔
(iv) «% error =» 8.2«%» ✔
Schemat punktowania
(i) 1 mark; do not accept a negative value. (ii) Award [2] for the correct final answer (accept 10.0–10.2 g). If 3.25 × 103 J is used, the answer is 11.7 g. (iii) Award [3] for the correct final answer (accept 0.31–0.32 g). If 7.80 g is used, the answer is 0.24 g. (iv) 1 mark; accept 8.1–8.6 %. If 7.80 g is used, the answer is 29.1 %.
Komentarz
Common traps: using a negative temperature change in (i), and forgetting that the uncertainty of a difference of two readings is the sum of their uncertainties (±0.4 K).