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Anhydrous question

.raiden.

Could someone explain how to do giii) ? Its not my working and I'm not sure if this person's method is the best way to go about it, thanks

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Reply 1

KanKan

Thats what I got too.

1. I worked out the molar mass of Na2S203 = 158.2
2. Then using the molar mass of the whole compound (248.2) - I divided it by 12.41 and got: 0.05
3. I multiplied 0.05 by 158.2 to get 7.91

Somehow I think I'm wrong too ... Is that the correct answer?

Reply 2

.raiden.

Original post by KanKan

No idea, no MS available for june 2012 papers, but why is the moles the same for anyhdrous, doesnt that mean less amount?

Reply 3

KanKan

Original post by raiden95

No idea, no MS available for june 2012 papers, but why is the moles the same for anyhdrous, doesnt that mean less amount?

Sorry, I don't understand your question. An anhydrous compound is just a compound without water - so you remove the moles of the H2O (5 X 18 = 90)

Reply 4

KanKan

Looking back at my work, I've done that question before, and I am correct :smile:

(teacher marked it)

Reply 5

.raiden.

Original post by KanKan

Sorry, I don't understand your question. An anhydrous compound is just a compound without water - so you remove the moles of the H2O (5 X 18 = 90)

I mean the moles, 90 is the molar mass, why is the moles the same when anhydrous is less??

Reply 6

KanKan

Original post by raiden95

I mean the moles, 90 is the molar mass, why is the moles the same when anhydrous is less??

The moles of the Na2S2O3 is the same, because it is not affected by the heat. Heating up the compound, removes the water but the salt is left behind - unaffected.