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Chlorine exists naturally as a mixture of two isotopes, 35Cl and 37Cl, in the abundance ratio of 3:1 - CIE - A-Level Chemistry - Question 1 - 2014 - Paper 1
Question 1
Chlorine exists naturally as a mixture of two isotopes, 35Cl and 37Cl, in the abundance ratio of 3:1. The mass spectrum of chlorine consists of five peaks. (i) Sugg... show full transcript
Worked Solution & Example Answer:Chlorine exists naturally as a mixture of two isotopes, 35Cl and 37Cl, in the abundance ratio of 3:1 - CIE - A-Level Chemistry - Question 1 - 2014 - Paper 1
Step 1
Suggest the mass numbers for these five peaks and the identities of the species responsible.
Answer
The mass numbers for the five peaks in the mass spectrum of chlorine are as follows:
- 35 - ³⁵Cl
- 37 - ³⁷Cl
- 70 - ³⁵Cl₂
- 72 - ³⁵Cl³
- 74 - ³⁷Cl₂
This accounts for the two isotopes of chlorine, each having different combinations of molecules.
Step 2
Predict the ratios of the abundances of the three species with the highest mass numbers.
Answer
The three species with the highest mass numbers are 74, 72, and 70. Given the abundance ratio of 35Cl to 37Cl is 3:1, we can say:
- The ratio of abundances for 37Cl will be 1,
- For 35Cl, it will be 3.
Hence, the ratio of abundances for peaks corresponding to 74, 72, and 70 will be approximately:
Step 3
Use the following data, together with relevant data from the Data Booklet, to calculate a value for the lattice energy of strontium chloride.
Answer
To determine the lattice energy of strontium chloride (SrCl₂), we can use the Born-Haber cycle. The formula for lattice energy is given by:
Where:
- (\Delta H_f = -830 : kJ : mol^{-1})
- (\Delta H_{atomization}(Sr) = +164 : kJ : mol^{-1})
- (\Delta H_{electron affinity} (Cl) = -349 : kJ : mol^{-1} \times 2) (for two Cl atoms)
Calculating:
- Lattice energy = -830 - (164 + 2(-349))
- Lattice energy = -830 - (164 - 698)
- Lattice energy = -830 - 164 + 698
- Lattice energy = -296 kJ mol⁻¹
Thus, the lattice energy for strontium chloride is approximately -296 kJ mol⁻¹.