Competition - post a photo you've taken of nature >>

Please help - difficult AS level q..

So the situation is that there are 2 identical beakers both with sodium thiosulphate in them; there are crosses on paper underneath them and basically as sulphur is added the crosses become decreasingly visible; here's the question:

Q) the cross disappears after 0.1 g of sulphur appear to make the solution cloudy. If the rate of reaction on the graph is measured in g/s calculate how long it tookthe cross to disappear at 0.50 mol. dm-3 sodium thiosulphate solution.

Any hints or help?

(edited 13 years ago)

Reply 1

hazbaz

Original post by Dededex

does it mention the volume used?. If so you could do Moles=conxv or moles=mass/rfm and then from that work out with a con of 0.5M

Reply 2

Dededex

Original post by hazbaz

does it mention the volume used?. If so you could do Moles=conxv or moles=mass/rfm and then from that work out with a con of 0.5M

Hmm no - I do have the equation : if I work out moles for sulphur then hence moles for sodium thiosulphate then can I use that concentration I get from c = nv to get the rate?

How do I get the rate? Is there a formula?

Reply 3

charco

Study Forum Helper

Original post by Dededex

Have you been given a graph by any chance?

Reply 4

Dededex

Original post by charco

Have you been given a graph by any chance?

Yes rate of reaction against concentration of sodium thiosulfate.

(edited 13 years ago)

Reply 5

mmmegan_x

work out how many moles 1g of sulfur is then look at the rate of reaction on the graph of that many moles of sodium thiosulphate - that's your rate of reaction.
next you work out how many g that 0.5 mol/dm3 sodium thiosulphate is using mass = mol x Mr
then you use the formula - rate of reaction = mass reacted/time taken - rearrange it for - time = mass reacted/rate of reaction - and slot in your values... so 79/whatever the rate of reaction was.
i think that's how you work it out... ha. i could be wrong. let me know what you get :biggrin:

(edited 13 years ago)

Reply 6

charco

Study Forum Helper

Original post by Dededex

Yes rate of reaction against concentration of sodium thiosulfate.

Then all you need to do is look at the graph...

Read across from 0.5 mol dm^-3 thiosulphate and that gives you the rate for that concentration.

The rate is measured in g/s so you now calculate to see how many of the s are needed to make 1.0g

Example:

If your rate is 0.1 g/s then it will take 10 s to make 1 g
If your rate is 0.2 g/s it will take 5 seconds to make 1 g

Reply 7

Dededex

Original post by charco

The graph doesn't go up to 0.50 mol. dm-3 = only to 0.15 also it's rate of reaction against concentration WTF

Reply 8

Dededex

Chemistry is the worst a level I swear!

Reply 9

charco

Study Forum Helper

Original post by Dededex

The graph doesn't go up to 0.50 mol. dm-3 = only to 0.15 also it's rate of reaction against concentration WTF

Yes, it's rate of reaction against concentration.

You are given the concentration, so you use the graph to read off the rate for the concentration that you need.

Are you sure that you don't want the rate at 0.05 mol dm^-3 ?

Check the question carefully...