cud anyone help me with this
mass of iron (II)ammonium sulphate used is 0.0693 +/- 1x10-4 g
Relative molar mass: 392.14
hence the molarity of standard solution:
............................. +/- ......................... moldm3
the volume of the standard solution is 1000cm3 by the way
how do i work out the errors with the molarity
2006-10-17 07:59:15 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
The molarity is worked out as normal then for the error you work out the molarity of the error 1 x 10-4 g in exactly the same way.
Alternatively you could work out the molarity of 0.0693 + 1 x 10-4g then subtract the molarity of 0.0693g from it. That's the long way round but if it makes more sense to you that way the answer will still be correct.
2006-10-17 09:03:35 · answer #1 · answered by Ellie 4 · 0⤊ 0⤋
Been a while since I did this but I think.....
Remember molarity = moles / volume.
1 mole of Ammonium Sulphate apparently weighs 392.14 g (according to your info).
You have 0.0693 g so (0.0693/392.14) moles.
Units of volume for this equation are litres ...1000cm3 = 1 litre
So equation becomes molarity = moles
You do the math.
To work out error do the same only instead of 0.0693 you are using 0.0001 g...
2006-10-17 08:38:30 · answer #2 · answered by Jericho Jack 1 · 0⤊ 0⤋
The error in the molarity is the error in the mass divided by the molar mass.
It certainly was a very accurate balance. Existed somewhere in the figment of a question-setter's imagination?
2006-10-17 20:58:17 · answer #3 · answered by Gervald F 7 · 0⤊ 0⤋
using the formula:- number of moles=mass in grams/RMM.
No. of moles=0.0693/392.14=0.000176 moles.
Using another formula, number of moles=molar conc.(in mol dm-3) x Volume,V of the substance(in dm-3), converting 1000cm3 to dm3 = 1dm3
0.000176=molar conc. x 1dm3
molar conc.=0.000176moldm-3
errors in molarity I'm not sure about, also this is a simple solution to your problem, just wanted to give you an indication on how to do it.
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2016-10-19 21:26:49 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Was just running through the maths, but have seen jericho jack answer and I agree with him. Seems right to me, but you do the sums and try to understand the logic.
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