[WORKED SOLUTION] CIE A Level Chemistry Paper 1 Summer 2019
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Course
Worked Solution Paper 1 Summer 2019 (9701)
Institution
CIE
Paper 1 A Level can be challenging and tricky since there's no clue regarding the answer in the mark scheme. It makes revision and learning process daunting.
This worked solution is made to help you tackling the paper 1 A Level Chemistry 9701 with step by step solution so that you can understand...
When the temperature is increased, the energy is being transferred to the gas sample. Thus, the number of
molecules with high energy increase. On the other hand, when temperature is increased, the peak of the
Boltzmann Distribution shifts to the right and lowered. Therefore the number of molecules with most
probable energy decrease.
Source: physical chemistry - Boltzmann factor vs. graph of
Maxwell–Boltzmann distribution - Chemistry Stack Exchange
,Find the relative atomic mass by using the percentage abundance of each isotopes and round the result.
112 × 3% + 114 × 56% + 115 × 41%
= 114.35
100%
Pay attention to the reactants.
NaHCO3 + H2SO4
H2SO4 is a dibasic acid (capable of releasing two H+), thus the reaction would be:
2NaHCO3 + H2SO4 → Na2SO4 + 2H2O + 2CO2
Tip: if you can’t figure out what would be the products of the reaction, just keep in mind that H2SO4 is a
dibasic acid while NaHCO3 doesn’t. This way, you can also know that the mol ratio between NaHCO3
and H2SO4 is 1:2.
Find the mol of H2SO4 that reacts with the washing powder. Don’t forget to convert the units.
7.15 × 10−3 × 0.1 = 7.15 × 10−4
Since the mol ratio between NaHCO3 and H2SO4 is 1:2
Relative molecular mass of NaHCO3 = 23 + 1 + 12 + 16 × 3 = 84 (Find the data of Mr in the data
booklet).
Find the mass by multiplying mol of reacting NaHCO3 and its Mr (Relative molecular mass)
1.43 × 10−3 × 84 = 0.12012
Find the percentage of mass by comparing the mass of NaHCO3 reacting with H2SO4 to the mass of the
washing powder, then multiply the result with 100% to get the percentage, then round the result.
, 0.12012
× 100% = 12.012%
1.00
10Ne = 1𝑠 2 2𝑠 2 2𝑝6
The shape of p orbital is similar to dumbbell and it has higher energy than the s orbital.
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