Zadanie 1b
Wzorowane na: Chemistry HL Paper 2, May 2022, TZ2, pytanie 1(b). 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-1d.
Sodium reacts vigorously with water to form an alkaline solution.
Polecenie
A 0.460 g piece of sodium was placed in 250.0 cm3 of water.
(i) Calculate the molar concentration of the sodium hydroxide solution formed. Assume that the volume of the solution is 250.0 cm3.
(ii) Calculate the volume of hydrogen gas produced, in cm3, if the temperature was 20.0 °C and the pressure was 102 kPa. Use sections 1 and 2 of the data booklet.
(iii) The volume of hydrogen collected in the experiment was smaller than the value calculated in (ii). Suggest one reason for this.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «n(Na) = 0.460/22.99 =» 0.0200 «mol» AND «n(NaOH) = n(Na)» ✔
«[NaOH] = 0.0200/0.2500 =» 0.0800 «mol dm−3» ✔
(ii) «» 0.0100 «mol» ✔
« so» V = 239 «cm3» ✔
(iii) the sodium had been partly oxidized «by air» / was coated with oxide or hydroxide, so less than 0.460 g of sodium metal reacted ✔
OR
some hydrogen escaped / ignited / dissolved before it was measured ✔
Schemat punktowania
(i) M1: amount of sodium (equal to the amount of NaOH); M2: concentration. Award [2] for a correct final answer. (ii) M1: amount of hydrogen from the 2:1 ratio; M2: volume from PV = nRT converted to cm3. Award [2] for a correct final answer; accept 238 to 240 cm3 (T = 293 K or 293.15 K). Award [1 max] for 0.239 given with the unit cm3 (no conversion from dm3). Award [1 max] for 478 cm3 (accept 477) from using 0.0200 mol as n(H2). (iii) [1] for any one valid reason; do not accept human error or the gas being collected too slowly.
Komentarz
Using n(Na) instead of n(H2) in PV = nRT doubles the volume (478 cm3), a frequent slip.