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Edexcel IAL Chemistry A2 6.P4 evaluate results and conclusions

Edexcel IAL Chemistry A2 6.P4 evaluate results and conclusions
Pearson Edexcel IAL Chemistry syllabusChemistry YCH11First assessment 2019

The two direct questions both test the same judgement: explain whether a lycopene value read from a graph is valid by referring to the range of plotted data.

How this is tested

  • Judge whether a graph-derived lycopene concentration is valid from the plotted data range.

Question 4(d)(ii)

[Maximum number: 1]

This question is about the nitration of methyl benzoate.

The equation for the reaction is shown.

Figure for Question 4(d)(ii) — Edexcel A-Level Chemistry A2

Procedure

Step 1 Weigh between 1.9 g and 2.1 g of methyl benzoate in a 50 cm350 \mathrm{~cm}^{3} conical flask.
Step 2 Slowly add 5 cm35 \mathrm{~cm}^{3} of concentrated sulfuric acid to the methyl benzoate with swirling and place the flask in an ice-water bath to cool.

Step 3 Place 2.0 cm32.0 \mathrm{~cm}^{3} of concentrated nitric acid into a test tube.
Cool the nitric acid by immersing the test tube in an ice-water bath before slowly adding 2.0 cm32.0 \mathrm{~cm}^{3} of concentrated sulfuric acid.
Allow this nitrating mixture to cool.
Step 4 Using a teat pipette, add the nitrating mixture very slowly to the conical flask, ensuring the temperature does not exceed 7C7^{\circ} \mathrm{C}.

Step 5 Allow the flask to stand at room temperature for about 15 minutes and then pour the contents into a beaker containing some crushed ice. Impure methyl 3-nitrobenzoate will form.

Step 6 Recrystallise the methyl 3-nitrobenzoate using methanol as the solvent.
Step 7 Weigh the dry crystals and determine their melting temperature.

1.95 g of methyl benzoate reacted with an excess of nitric acid to form
1.51 g of methyl 3-nitrobenzoate.
[Molar mass values: methyl benzoate, C6H5CO2CH3=136 g mol1\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{CH}_{3}=136 \mathrm{~g} \mathrm{~mol}^{-1}
methyl 3-nitrobenzoate, C6H4CO2CH3NO2=181 g mol1\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{CH}_{3} \mathrm{NO}_{2}=181 \mathrm{~g} \mathrm{~mol}^{-1} ]

Give one possible reason why the yield in (d)(i) is less than 100\%.