The first six ionisation energies of an element, M, in kJ mol−1 are 660, 1267, 2218, 3313, 7863 and 9500. M forms its only chloride by heating M with chlorine. What is the equation for the formation of the chloride of M?
Source: SJChO Mock 3 P1 Q3
The relative abundances of all the isotopes present in a sample of nickel are shown.
What is the relative atomic mass of nickel calculated from these data?
SJChO Mock 3 P1 Q2
The relative atomic mass of antimony, which consists of the isotopes 121Sb and 123Sb, is 121.8. What is the percentage of 123Sb in the isotopic mixture?
2019 RVHS J1 Common Test P1 Q2
If there are n molecules in 10 g of methane, CH4, how many molecules are in 55 g of carbon dioxide?
2023 SJChO P1 Q39
Which of the following statements about 20.3 g of Co2(SO4)3 is false? [Molar mass = 406 g mol−1]
2019 RVHS J1 Common Test P1 Q1
Which of the following quantities is equal to the Avogadro constant?
2021 EJC J2 Prelim P1 Q8
Which substances contain the same number of atoms?
1 1.20 mol of NH3
2 1.445 × 1024 molecules of O2
3 36.0 g of C6H12O6
4 57.6 dm3 of Cl2 at room temperature and pressure
2024 VS O Prelim P1 Q10
Complete combustion of a sample of a hydrocarbon, Y, gave 0.66 g of carbon dioxide and 0.36 g of water. What is the empirical formula of Y?
2020 YIJC J2 Prelim P1 Q20
A 2.0 g sample containing 0.0130 mol of an organoboron compound, with the formula CxHyBOz, was burnt in an excess of oxygen. 4.58 g of CO2 and 1.76 g of H2O were produced. What is the molecular formula of the organoboron compound?
2021 SAJC J2 Prelim P1 Q3
The empirical formula of an oxide of element X is X2O5. 106.5 g of X2O5 contains 60 g of oxygen. What is the identity of X?
2025 CHIJ O Prelim P1 Q12
Sodium hydrogencarbonate, NaHCO3, can be prepared from sodium sulfide, Na2S, by the following reactions.
Na2S(s) + CaCO3(s) → CaS(s) + Na2CO3(s)
Na2CO3(s) + H2O(l) + CO2(g) → 2NaHCO3(s)
Assuming that the percentage yield in each step is identical, 20.5 kg of sodium hydrogencarbonate (Mr = 84.0) was formed from 50 kg of sodium sulfide (Mr = 78.1). What is the percentage yield of each step?
2022 VJC J1 Promo P1 Q7
What is the volume of distilled water that must be added to 50 cm3 of 0.280 mol dm−3 KMnO4 to dilute it to 0.160 mol dm−3?
2022 AMSR Topical Revision Q10
Amedeo Avogadro discovered that equal volumes of gases at the same temperature and pressure contain the same number of molecules. Two closed containers, A and B, hold gases at the same temperature and pressure. The mass of gas A is equal to the mass of gas B, but gas A has a higher density than gas B. Which of the following statements is true?
2023 SJChO P1 Q42
Which graph is a correct representation of Boyle’s law?
SJChO Mock 3 P1 Q27
0.0353 g of a sample of volatile organic compound was completely vaporised at 50 kPa and 300 K. The vapour occupied a volume of 40 cm3. On complete combustion in excess oxygen, 80 cm3 of carbon dioxide and 80 cm3 of water vapour were formed. All gaseous volumes were measured under identical conditions. What would be the molecular formula of the compound? (R = 8.31 J K−1 mol−1)
2022 NJC J2 Prelim P1 Q4
In an experiment, 0.100 g of a volatile liquid Q formed 0.0250 dm3 of vapour at 100 °C and 1 bar. What is the relative molecular mass of Q?
2022 YIJC J2 Prelim P1 Q6
Deuterium, 2H, is an isotope of hydrogen. In an experiment, 18.0 g of gas A at 40 °C is found to occupy the same volume as 0.6 g of deuterium gas, 2H2, at 20 °C. The pressure of gas A is twice that of the deuterium gas. What is the relative molecular mass of gas A?
2024 VJC J2 Prelim P1 Q5
A 20 dm3 vessel contains 0.25 mol of ammonia gas and a 30 dm3 vessel contains 0.42 mol of hydrogen chloride gas. When the two vessels were connected at 500 K, the following reaction occurs: NH3(g) + HCl(g) → NH4Cl(s). What is the resultant pressure, in Pa, in the combined vessel? (R = 8.31 J K−1 mol−1)
2023 TJC J2 Prelim P1 Q8
An equimolar mixture of H2 and O2 was placed in a sealed, rigid container at room temperature. The total pressure inside the container is P Pa. The mixture was sparked, and H2 and O2 reacted until H2 was completely consumed. What is the partial pressure, in Pa, of O2 in the container after the reaction at room temperature?
2020 RI J1 Promo P1-P3 Q6
A gaseous sample containing x dm3 of gaseous methanol, CH3OH, was burnt in 4x dm3 of oxygen to produce carbon dioxide and water vapour. The total pressure in the system after combustion was P. Which statements are correct?
1 Methanol reacts with oxygen in a 2 : 3 mole ratio.
2 The partial pressure of carbon dioxide in the system is 0.18P.
3 Assuming these gases behave ideally, they occupy an insignificant volume.
2021 DHS J2 Prelim P1 Q5
F2(g) and Xe(g) are added to a container of fixed volume. The initial partial pressure of F2(g) is 7.0 atm and that of Xe(g) is 1.6 atm. When all of the Xe(g) has reacted with F2(g) to form a solid compound, C, the pressure of unreacted F2(g) is 3.8 atm. The temperature remained constant throughout. What is the formula of the solid compound C?
2022 Gaseous State Topical Revision Q12
Analysis of a mixture of sulfur-containing gases shows that hydrogen sulfide, H2S, and carbon disulfide, CS2, are present in a 2 : 1 mole ratio. The mixture is burned in excess oxygen, forming SO2, CO2 and H2O. Which row describes the SO2 : CO2 mole ratio after complete combustion, and the relative deviation from ideal behaviour of the gases CO2 and CS2?
| SO2 : CO2 mole ratio | relative deviation | |
|---|---|---|
| A | 3 : 1 | CS2 deviates more than CO2 |
| B | 4 : 1 | CS2 deviates more than CO2 |
| C | 3 : 1 | CO2 deviates more than CS2 |
| D | 4 : 1 | CO2 deviates more than CS2 |
2024 DHS J2 Prelim P1 Q3
CM + 2Cl2 → MCl4[1]
The big jump is after the 4th ionisation energy (3313 to 7863), so M has 4 valence electrons and is in Group 14.
M bonds to four chlorine atoms, giving MCl4. One mole of M needs 2 mol of Cl2.
A58.8[1]
The abundances add up to 69.3 + 26.7 + 1.20 + 3.80 + 1.00 = 102, so divide by 102, not 100.
Ar = (58 × 69.3 + 60 × 26.7 + 61 × 1.20 + 62 × 3.80 + 64 × 1.00) ÷ 102 = 5994 ÷ 102 = 58.8
A40%[1]
Let 121Sb be x%: 121x + 123(100 − x) = 12 180, so 12 300 − 2x = 12 180 and x = 60.
123Sb is 100 − 60 = 40%. Option D is the percentage of 121Sb.
B2n[1]
10 g of CH4 (Mr 16) is 0.625 mol, which is n molecules.
55 g of CO2 (Mr 44) is 1.25 mol, twice as many moles, so it has 2n molecules.
DIt contains 3.01 × 1022 ions.[1]
n = 20.3 ÷ 406 = 0.0500 mol. Each unit has 2 Co3+ and 3 SO42−, so there are 0.250 mol of ions, 1.51 × 1023, not 3.01 × 1022.
A: 0.0500 mol of gas at 293 K and 1 atm is 0.0500 × 8.31 × 293 ÷ 101 325 = 1.2 × 10−3 m3. B: 12 × 16.0 ÷ 406 = 47.3%. C: 3 × 0.0500 = 0.15 mol.
CThe number of ions in 68.05 g of CaSO4.[1]
CaSO4 has Mr = 40.1 + 32.1 + 4 × 16.0 = 136.2, so 68.05 g is 0.50 mol. Each unit is 2 ions, Ca2+ and SO42−, so there is 1.0 mol of ions.
A is 1 mol of O2, which is 2 mol of atoms. B is pV/RT = 0.067 mol. D is 1 mol of C2H6, which has 2 mol of C atoms.
D1, 2, 3 and 4[1]
1: 1.20 × 4 = 4.80 mol of atoms. 2: 1.445 × 1024 ÷ 6.02 × 1023 = 2.40 mol of O2, which is 4.80 mol of atoms.
3: 36.0 ÷ 180 = 0.200 mol, and 0.200 × 24 atoms = 4.80 mol. 4: 57.6 ÷ 24 = 2.40 mol of Cl2, which is 4.80 mol of atoms. All four are equal.
CC3H8[1]
nC = nCO₂ = 0.66 ÷ 44 = 0.015 mol. nH = 2 × 0.36 ÷ 18 = 0.040 mol.
C : H = 0.015 : 0.040 = 3 : 8, so the empirical formula is C3H8.
CC8H15BO2[1]
nC = 4.58 ÷ 44.0 = 0.104 mol, 8 per molecule (÷ 0.0130). nH = 2 × 1.76 ÷ 18.0 = 0.196 mol, 15 per molecule.
Mr = 2.0 ÷ 0.0130 = 154. C8H15B accounts for 96 + 15 + 10.8 = 121.8, leaving 32 for 2 O atoms: C8H15BO2.
Aphosphorus[1]
Mass of X = 106.5 − 60 = 46.5 g. nO = 60 ÷ 16 = 3.75 mol, so nX = 3.75 × 2/5 = 1.50 mol.
Ar(X) = 46.5 ÷ 1.50 = 31, which is phosphorus.
B43.7%[1]
50 000 g of Na2S is 640 mol, which could give 2 × 640 = 1280 mol of NaHCO3. Only 20 500 ÷ 84.0 = 244 mol formed, an overall yield of 19.1%.
With the same yield y in both steps, y × y = 0.191, so y = 0.437 = 43.7%. Option C forgets that each Na2CO3 gives 2NaHCO3.
B37.5 cm3[1]
c1V1 = c2V2: 0.280 × 50 = 0.160 × V2, so V2 = 87.5 cm3.
V2 is the final volume, so the water added is 87.5 − 50 = 37.5 cm3. Option D is the final volume.
CThe number of molecules in A < the number of molecules in B[1]
The masses are equal but A is denser, so A occupies the smaller volume (D is false).
At the same temperature and pressure a smaller volume holds fewer molecules, so A has fewer molecules than B. With the same mass, its molar mass is larger (A is false). The number of atoms depends on the formulae, which are not given.
Agraph A[1]
Boyle's law: at constant temperature p ∝ 1/V, so p against 1/V is a straight line through the origin.
p against V is a curve, not a straight line. C is Charles's law, V ∝ T, and V against 1/T matches neither law.
BC2H4O[1]
n = pV/RT = (50 × 103 × 40 × 10−6) ÷ (8.31 × 300) = 8.02 × 10−4 mol, so Mr = 0.0353 ÷ 8.02 × 10−4 = 44.0.
Volume ratio = mole ratio: 40 cm3 gives 80 cm3 of CO2 (2 C) and 80 cm3 of H2O (4 H). C2H4 is 28, so one O makes 44: C2H4O.
D0.100 × 373 × 22.70.0250 × 273[1]
At s.t.p. (273 K and 1 bar) the vapour would occupy 0.0250 × 273/373 dm3, since V ∝ T at constant pressure.
n = that volume ÷ 22.7, and Mr = 0.100 ÷ n = (0.100 × 373 × 22.7) ÷ (0.0250 × 273).
C64.1[1]
2H2 has Mr = 4.0, so n = 0.6 ÷ 4.0 = 0.15 mol. With the same V, n ∝ p/T: nA = 0.15 × 2 × 293/313 = 0.281 mol.
Mr(A) = 18.0 ÷ 0.281 = 64.1
A14 127[1]
NH3 is limiting: 0.25 mol reacts with 0.25 mol of HCl, leaving 0.42 − 0.25 = 0.17 mol of HCl. NH4Cl is a solid, so it adds nothing to the pressure.
p = nRT/V = 0.17 × 8.31 × 500 ÷ (50 × 10−3) = 14 127 Pa.
D0.25P[1]
Equal amounts, so each gas starts at 0.5P. 2H2 + O2 → 2H2O: the H2 (0.5P) uses half as much O2, 0.25P.
0.5P − 0.25P = 0.25P of O2 is left. The water is liquid at room temperature, so it adds no pressure.
B1 and 2 only[1]
2CH3OH + 3O2 → 2CO2 + 4H2O, so 1 is correct. x dm3 of methanol uses 1.5x of O2, leaving 2.5x, and makes x of CO2 and 2x of steam: 5.5x in all.
p(CO2) = x ÷ 5.5x × P = 0.18P, so 2 is correct. 3 is wrong: an ideal gas's particles have negligible volume, but the gas itself fills its container.
DXeF4[1]
At constant volume and temperature, partial pressure is proportional to amount. Xe used: 1.6 atm. F2 used: 7.0 − 3.8 = 3.2 atm.
Xe : F2 = 1.6 : 3.2 = 1 : 2, so Xe + 2F2 → XeF4.
Brow B[1]
2H2S + 3O2 → 2SO2 + 2H2O and CS2 + 3O2 → CO2 + 2SO2. 2 mol of H2S and 1 mol of CS2 give 4 mol of SO2 and 1 mol of CO2.
CS2 has more electrons than CO2, so its instantaneous dipole–induced dipole forces are stronger and its molecules larger: it deviates more.