// StartMathJax Script window.MathJax = {loader: {load: [ 'input/asciimath', 'ui/lazy', 'output/chtml', 'ui/menu']} }; (function() { var script = document.createElement('script'); script.src = "https://cdn.jsdelivr.net/npm/mathjax@3/es5/startup.js"; script.async = true; document.head.appendChild(script); })(); ---------- (Different files) ---------- // UpdateTypeset Script config = { attributes: true, childList: true, subtree: true }; // Callback function to execute when mutations are observed callback = (mutationList, observer) => { for (mutation of mutationList) { if (mutation.type === 'childList') { console.log('A child node has been added or removed.'); MathJax.typeset(); } else if (mutation.type === 'attributes') { console.log(`The ${mutation.attributeName} attribute was modified.`); } } }; // Create an observer instance linked to the callback function observer = new MutationObserver(callback); document.onreadystatechange = () => { if (document.readyState === 'complete') { console.log("Loaded fully according to readyState") targetNode = document.getElementById('content-wrapper') console.log(targetNode) // Start observing the target node for configured mutations observer.observe(targetNode, config); } }
top of page

Experimental Considerations: Past Paper Question Analysis

Updated: Jul 23, 2023

P3 June 2017 Q3




This question is about practical considerations during experimentation.




If you haven't gone through the paper, please look up the questions at www.aqa.org.uk, or click the direct link below, and have a go...




This question is in 9 parts with 18 marks available


Question 3

Anti-bumping granules prevent the formation of large bubbles - it reduces them but DOES NOT prevent them.


Couple of ways to do this - here's one...

Mass of benzoate = 1.05 x 5.0 = 5.25g and moles of benzoate = 5.25 / 150.0 = 0.0350 mol. Moles of hydroxide = (30 x 2) ÷ 1000 = 0.06 mol.

∴ NaOH is in excess.

Ensures the complete hydrolysis of the sodium benzoate.


The ester and ethanol are flammable (many organic substances are)


Refluxing allows vapours to condense and return to the mixture.


C6H5COONa + HCl → C6H5COOH + NaCl


Sodium benzoate is soluble because it is ionic, forming C6H5COO– and Na+ Benzoic acid is insoluble because the benzene ring is non-polar (even though the COOH group is).


Dissolve crude substance in a hot solvent.

The resulting solution should be of minimum volume Filter the hot solution to remove insoluble impurities

Cool to recrystallise Filter under reduced pressure (Buchner funnel)

Wash, with cold solvent, and dry


Moles of benzoic acid = 5.12 ÷122 = 0.042 mol)

Percentage yield = (0.042 ÷ 0.04) x 100 = 105%

More product than expected so perhaps not completely dry, or there are impurities present.


 

Please share your thoughts through the comments box below - I look forward to hearing from you.



 






コメント

5つ星のうち0と評価されています。
まだ評価がありません

評価を追加
bottom of page