IB Chemistry HL 1.4 Emission Spectra Question Bank
Practise IB Chemistry HL 1.4 with evidence-led questions on emission spectra, energy transitions and electron configurations.
- Syllabus
- First assessment 2025
- Course
- Chemistry HL
- Level
- HL
Practise IB Chemistry HL 1.4 with evidence-led questions on emission spectra, energy transitions and electron configurations.
How much ethanol contains 1.20×1024 atoms of carbon?
Avogadro's constant, L or NA:6.02×1023 mol−1
0.333 mol
0.500 mol
1.00 mol
2.00 mol
C
Two groups of students (Group A and Group B) carried out a project* on the chemistry of some group 7 elements (the halogens) and their compounds.
The students reacted ICl(l) with CsBr(s) to form a yellow solid, CsICl2( s), as one of the products. CsICl2( s) has been found to produce very pure CsCl(s) which is used in cancer treatment.
To confirm the composition of the yellow solid, Group A determined the amount of iodine in 0.2015 g of CsICl2( s) by titrating it with 0.0500moldm−3Na2 S2O3(aq). The following data were recorded for the titration.

Calculate the percentage of iodine by mass in CsICl2( s), correct to three significant figures.
(330.71126.90×100)=38.4%;
Determine the amount, in mol, of 0.0500moldm−3Na2 S2O3(aq) added in the titration.
(100024.20×5.00×10−2)=1.21×10−3/0.00121( mol);
Calculate the mass of iodine, in g , present in the sample of CsICl2( s).
(126.90×6.05×10−4)=7.68×10−2/0.0768( g);
Marking guidance:
Accept alternate method e.g. (6.10×10−4×126.9) or (0.2015×0.384)=7.74×10−2/0.00774 (g).
Determine the percentage by mass of iodine in the sample of CsICl2( s), correct to three significant figures, using your answer from (v).
(0.20157.68×10−2×100)=38.1%;
Answer must be given to three significant figures.
Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula PH3.
2.478 g of white phosphorus was used to make phosphine according to the equation:
Calculate the amount, in mol, of white phosphorus used.
« ⟨4×30.972.478⟩ » =0.02000 «mol»
Determine the volume of phosphine, measured in cm3 at standard temperature and pressure, that was produced.
< 22.7 x 1000 x 0.02000 >=454 < cm^3 >
Marking guidance:
Accept methods employing p V=n R T, with p as either 100(454 cm^3) or 101.3 kPa(448 cm^3). Do not accept answers in dm^3.
Impurities cause phosphine to ignite spontaneously in air to form an oxide of phosphorus and water.
The oxide formed in the reaction with air contains 43.6% phosphorus by mass. Determine the empirical formula of the oxide, showing your method.
n(P)=30.9743.6=1.41 moln(O)=16.00100−43.6=3.53 moln(P)n(O)=1.413.53=2.50
Empirical formula:
P2O5
Accept other methods where the working is shown.
The molar mass of the oxide is approximately 285 g mol−1.
Determine the molecular formula of the oxide.
« 141.9285=2.00, so molecular formula =2×P2O5= » P4O10