Maths question OCR A Chemistry HELP NEEDED!!

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Sidd1
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For this question, when they said "10% increase in KI" I assumed the volume increases by 10% turns out its the concentration..... why is that?
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Harrybeld
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(Original post by Sidd1)
For this question, when they said "10% increase in KI" I assumed the volume increases by 10% turns out its the concentration..... why is that?
Maybe it's related to a previous part of the question? This would make sense if the practical had a control volume of 50cm3.
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bfm.mcdermott
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(Original post by Sidd1)
For this question, when they said "10% increase in KI" I assumed the volume increases by 10% turns out its the concentration..... why is that?
I read it as - the volume is 50cm3 but there are 10% more moles of KI.
Since there is a fixed volume, an increase in moles means an increase in concentration.

Also, when you are working about how much of a substance (KI) is needed to react with other substance (Pb(NO3)2), you work it out using moles, so I would do the 10% in moles too, if that makes sense.
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Sidd1
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(Original post by Harrybeld)
Maybe it's related to a previous part of the question? This would make sense if the practical had a control volume of 50cm3.
Yeah it is but the previous part isn't really needed here since they give all the information required again including the equation.
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Sidd1
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(Original post by bfm.mcdermott)
I read it as - the volume is 50cm3 but there are 10% more moles of KI.
Since there is a fixed volume, an increase in moles means an increase in concentration.

Also, when you are working about how much of a substance (KI) is needed to react with other substance (Pb(NO3)2), you work it out using moles, so I would do the 10% in moles too, if that makes sense.
Okay so I get how increasing moles increases concentration. However, how do you know the volume is fixed and that's not affected? *I'm so confused*
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gongoozled goose
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(Original post by Sidd1)
Okay so I get how increasing moles increases concentration. However, how do you know the volume is fixed and that's not affected? *I'm so confused*
The question never says about increase in volume...if the volume is changed, it would be said so.
Last edited by gongoozled goose; 7 months ago
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Schoolboy101
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(Original post by Sidd1)
Okay so I get how increasing moles increases concentration. However, how do you know the volume is fixed and that's not affected? *I'm so confused*
Because it said there is 10% more KI. Have 10% increased volume of KI isn’t actually more KI if you think about because the moles is exactly the same just the solvent has increased in amounts. To have more of X, in this case KI, you need to increase how much you physically have it so more moles.
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Sidd1
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(Original post by lyer_in_hellfyre)
The question never says about increase in volume...if the volume is changed, it would be said so.
Oh okay so this means concentration is affected and therefore moles too?
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Sidd1
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(Original post by Schoolboy101)
Because it said there is 10% more KI. Have 10% increased volume of KI isn’t actually more KI if you think about because the moles is exactly the same just the solvent has increased in amounts. To have more of X, in this case KI, you need to increase how much you physically have it so more moles.
Ohhhh right that makes so much sense! Thank you
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paulkisho
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What's the answer?
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Sidd1
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(Original post by paulkisho)
What's the answer?
The answer is 3.3 mol dm^-3
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paulkisho
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It's asking for concentration of KI after the 10% extra is used
So work out moles of the Pb(NO3)2 and multiply by 2 (Ratios). So now you have moles of KI, then to work out concentration, take moles of KI and multiply by 1000/50 (n=c x v) to get concentration of KI. Find 10% of this concentration and add it on to this concentration.

So this is what I did.
n(Pb(NO3)2) = 1.50 * (50/1000) = 0.075 mol
n(KI)= 0/075 * 2 = 0.15 mol
n=cv, therefore to workout conc(KI) = 0.15/(50/1000) = 3
To get the 10% extra conc of KI : 3 * (10/100) = 0.3
3+0.3= 3.3 moldm-3 which is the concentration of KI used

Hope that helps
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Sidd1
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(Original post by paulkisho)
It's asking for concentration of KI after the 10% extra is used
So work out moles of the Pb(NO3)2 and multiply by 2 (Ratios). So now you have moles of KI, then to work out concentration, take moles of KI and multiply by 1000/50 (n=c x v) to get concentration of KI. Find 10% of this concentration and add it on to this concentration.

So this is what I did.
n(Pb(NO3)2) = 1.50 * (50/1000) = 0.075 mol
n(KI)= 0/075 * 2 = 0.15 mol
n=cv, therefore to workout conc(KI) = 0.15/(50/1000) = 3
To get the 10% extra conc of KI : 3 * (10/100) = 0.3
3+0.3= 3.3 moldm-3 which is the concentration of KI used

Hope that helps
Thank you, I get this now
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Sidd1
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How is everyone finding chapter 24 - transition elements? I find that chapter quite hard like the cis-trans isomerism drawing and in general stereoisomerism. If anyone finds this chapter okay any advice on how to tackle it in terms of resources and questions?

Thanks!
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