9701/52

Chemistry 9701/52February/March 2022

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

2
questions
30
marks
75
minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q1PlanningAnalysis, Conclusions and EvaluationFree sample

A student has a sample of copper(II) sulfate crystals, CuSO4xH2O\text{CuSO}_4\boldsymbol{\cdot}x\text{H}_2\text{O}. The student wants to show that the value of xx is 5.

The student uses the following method.

step 1 Weigh a clean crucible on a balance reading to two decimal places. Record the mass.

step 2 Place the sample of CuSO4xH2O\text{CuSO}_4\boldsymbol{\cdot}x\text{H}_2\text{O} into the crucible. Record the mass.

step 3 Heat the crucible gently for about 1 minute then strongly for about 4 minutes.

step 4 Weigh the crucible and contents. Record the mass.

(a)

Identify the instruction that is missing between step 3 and step 4.

1M
(b)

Explain why gentle heating takes place in step 3.

1M
(c)

Name the apparatus that should be used to hold the crucible during heating.

1M
(d)

The method is incomplete.

State the step(s) that should be carried out to complete the method.

1M
(e)

The student records their results in Table 1.1.

Table 1.1

mass/g\text{mass}/\text{g}
mass of crucible13.60
mass of crucible + contents before heating21.09
mass of crucible + contents at the end of experiment17.94
(i)

Calculate the experimental value of xx from these results.

experimental value of x=..............................\text{experimental value of } x = \text{..............................}
3M
(ii)

Suggest why the experimental value of xx varies from the expected value of 5.

If you were unable to obtain an answer to (e)(i), use the experimental value x=6.9x = 6.9. This is not the correct answer.

1M
(f)

The empty crucible weighs 13.60 g.

Calculate the percentage error in this measurement.

Show your working.

percentage error=..............................\text{percentage error} = \text{..............................}
1M

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