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2009 HSC Question 14

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bloop:
Hi,

I found this question from the 2009 HSC and it is a multiple choice question. I'm struggling to figure this out and any help would be great :D

MisterNeo:

--- Quote from: bloop on July 21, 2017, 11:35:24 pm ---Hi,

I found this question from the 2009 HSC and it is a multiple choice question. I'm struggling to figure this out and any help would be great :D

--- End quote ---

Hey! :)
First off, you should write out the chemical equation.

It says that 29.5mL of NaOH standard solution was required to neutralise the acid. So you find the moles of NaOH, then use stoichiometry to find moles of citric acid.
0.0295L x 0.55mol/L = 0.016225mol
Use stoichiometry, 1:3 ratio.
0.016225mol /3 = 0.00541mol
Then find how many grams of acid this is using the molar mass given.
0.00541mol x 192.12g/mol = 1.039g
Finally, it wants it in g/L so we divide the grams of acid by how much acid was used in the titration.
1.039g / 0.025L = 41.56g/L
Thus, the answer is B (41.6g/L).
Hope this helps ;D

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