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How would I work this out

I'm don't understand how to work out this question. It's on physics and maths tutor A level chemistry question 11 in the Amount of a substance QP.
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(edited 12 months ago)

Reply 1

Original post by Slugzie(:
I'm don't understand how to work out this question. It's on physics and maths tutor A level chemistry question 11 in the Amount of a substance QP.
Screen Shot 2024-05-18 at 11.45.06.png

Chem A2 - page 101.png
Heres what I did so pretty much you start from the titration by working out the moles of the sodium thiosulfate then you can just divide by two to get the moles of I2. You can link the first and second equation because the only place where iodine is supplied from is the first equation. Therefore all of the moles that are reacting with the sodium thiosulfate must have came from the first reaction. Once you have the moles of I2, you can see thats its in a 1:1 ratio to ClO- so that lets you get the moles for NaClO. From there work out its mass so we have a 25 cm3 sample from the 100 cm3. We multiply by four to get back to the number of moles in 100 cm3. The reason we can do this is because the 100cm3 comes from the orginal diluted sample so it will still have all the moles from that intial 10cm3 present, just a lower concentration. This is not the same with the 25cm3 sample where it will actually have a lower number of moles. From here we now just take our mass of NaClO and divide it by the mass of the entire bleach solution which we can derive from density* volume. We get the answer to be 4.16%

Reply 2

Original post by Peanut11200
Chem A2 - page 101.png
Heres what I did so pretty much you start from the titration by working out the moles of the sodium thiosulfate then you can just divide by two to get the moles of I2. You can link the first and second equation because the only place where iodine is supplied from is the first equation. Therefore all of the moles that are reacting with the sodium thiosulfate must have came from the first reaction. Once you have the moles of I2, you can see thats its in a 1:1 ratio to ClO- so that lets you get the moles for NaClO. From there work out its mass so we have a 25 cm3 sample from the 100 cm3. We multiply by four to get back to the number of moles in 100 cm3. The reason we can do this is because the 100cm3 comes from the orginal diluted sample so it will still have all the moles from that intial 10cm3 present, just a lower concentration. This is not the same with the 25cm3 sample where it will actually have a lower number of moles. From here we now just take our mass of NaClO and divide it by the mass of the entire bleach solution which we can derive from density* volume. We get the answer to be 4.16%

Thank you so much! Do you have any tips in general for answering these types of questions in the amount of a substance topic?

Reply 3

Original post by Slugzie(:
Thank you so much! Do you have any tips in general for answering these types of questions in the amount of a substance topic?

Usually start by working out the moles from the information they give you then follow what the rest of the question says. The pattern most of the time is work out moles from info, adjust moles via the ratios then if its a titration multiply the sample size by a factor to get it to the amount of moles in the original solution. That's what I did above.

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