ATAR Notes: Forum
HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Chemistry => Topic started by: 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
-
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
Hey! :)
First off, you should write out the chemical equation.
}+\ce{3NaOH(aq)}\rightarrow\ce{Na3C6H5O7(aq)}+\ce{3H2O(l)})
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