IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: ksitna on January 12, 2010, 04:04:38 pm

Title: need help with chem AS
Post by: ksitna on January 12, 2010, 04:04:38 pm
17.1 g of aluminium sulfate, Al2(SO4)3, was dissolved in water.
Calculate the number of sulfate ions, SO4
2–, present in the solution formed.
[Assume the molar mass of Al2(SO4)3 is 342 g mol–1 and the Avogadro Constant is
6 × 1023 mol–1.]

how do u do this
can u help me pleassee
Title: Re: need help with chem AS
Post by: 01chris on January 12, 2010, 06:13:50 pm
The formula of the compound Al2(SO4)3 tells you that you have 3 SO4 ions for every molecule of the compound.

You need to work out how many moles of the compound you have: n = m/M  -or as I like to remember it- no. moles = grams/RAM.

17.1g/342g mol-1 = 0.05mol.

Avagadro's number says that a mole of a substance = 6x1023 particles. You have 0.05 mol therefore 0.05 x 6x1023 particles (which equals 3x1022 ).

Going back to what we said at the start: for every Al2(SO4)3 molecule you have 3 SO4 ions. You have 3x1022 Al2(SO4)3 therefore you have 3 x 3x1022 SO4 ions (= 9x1022 ).

Does that help?
Title: Re: need help with chem AS
Post by: Dento on January 12, 2010, 06:16:59 pm
hey ksitna,
here's how to deal with these questions,

first, find the numer of moles of the sulphate ions which is >>>> 17.1/342 = 0.05 moles

then multply this mole number with the avogadro's constant and 3.    ( 3 is the number of sulphates given in the question >> Al2(SO4)"3" )

so, it goes this way,



moles of sulphate = 0.05 moles
number of sulphate ions present = 0.05 x (6x10 raised 23) x 3
= 9x10 raised 22
Answer is "D"

this is the final answer, hope i helped u, if there is anything else, feel free to ask
good luck this thursday !!
Title: Re: need help with chem AS
Post by: ksitna on January 12, 2010, 06:34:56 pm
THANK UUU!!!!!!!
I OWE U GUYS ONE
THANKS A LOOTTTT ;D
Title: Re: need help with chem AS
Post by: ksitna on January 12, 2010, 06:37:02 pm
sooo sorry to be a bother
AGAIN
but um are the oxidation stuff and flame tests required for this module?
Title: Re: need help with chem AS
Post by: 01chris on January 12, 2010, 06:49:04 pm
No worries! Glad I could help.

I'm afraid I can't answer your other question - I'm doing a different board (OCR). Sorry
Title: Re: need help with chem AS
Post by: ksitna on January 12, 2010, 06:51:13 pm
Quote from: 01chris on January 12, 2010, 06:49:04 pm
No worries! Glad I could help.

I'm afraid I can't answer your other question - I'm doing a different board (OCR). Sorry


ohh!!
no prob
thanks anywayy :D
really appreciated it
Title: Re: need help with chem AS
Post by: mountaindewaddict on January 12, 2010, 06:54:51 pm
No flame tests or oxidation in unit 1.
Title: Re: need help with chem AS
Post by: sweetie on February 21, 2010, 07:38:57 pm
Hey guyz,
 doubt in organic chemistry P1

Q1) which compound is unaffected by hot alkaline potassium manganate (VII) and gives hydrogen when treated with sodium?
a.(CH3)2CHCOCH3
b.(CH3)3COH
c.CH3CH2CH(OH)CH3
d.CH3CO2CH(CH3)2

I thought da answer was C but it is wrong  :'( dunno why!!!!!

I really appreciate any help from u guyz  :-*
Title: Re: need help with chem AS
Post by: nid404 on February 21, 2010, 08:23:29 pm
See b is a primary alcohol which can get oxidised, c is a secondary alcohol which can also get oxidised
so it leaves us with a and d
a is a ketone, which won't be affected by KMNO4...it won't get oxidised further
so it leaves us with d....
By the way what does the ms say?...cuz i keep forgetting reaction mechanisms...so i may have messed up
Title: Re: need help with chem AS
Post by: sweetie on February 22, 2010, 05:06:26 pm
ya i guess u messed up a bit , the ms says "B"
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 05:44:43 pm
yaa....stupid me....I jst realized...

sorry
B is actually 2 methyl, propan-2-ol
This won't be affected by KMNO4 because it a tertiary alcohol....
alcohols=water in organic chem....so it will give hydrogen with sodium too
Title: Re: need help with chem AS
Post by: sweetie on February 22, 2010, 05:52:12 pm
why is it not "c"
It is also an alcohol and no double bond, so no reaction with KMnO4?
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 05:55:21 pm
Quote from: notty_me on February 22, 2010, 05:52:12 pm
why is it not "c"
It is also an alcohol and no double bond, so no reaction with KMnO4?


C is a secondary alcohol...it will form a ketone when oxidised

unfortunately i'd mistaken B for a primary alcohol...not noticing it was (CH3)3
Title: Re: need help with chem AS
Post by: sweetie on February 22, 2010, 06:07:08 pm
got it  ;D

thanx  :-*
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 06:08:15 pm
Quote from: notty_me on February 22, 2010, 06:07:08 pm
got it  ;D

thanx  :-*

lol...anytime
Title: Re: need help with chem AS
Post by: sweetie on February 22, 2010, 07:03:45 pm
me again ;D

can u tell me what r da products produced on reforming hexane?
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 07:08:27 pm
benzene and hydrogen
Title: Re: need help with chem AS
Post by: sweetie on February 22, 2010, 07:14:32 pm
cud it be cyclohexane and hydrogen
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 07:20:36 pm
Reforming takes straight chain hydrocarbons in the C6 to C8 range from the gasoline or naphtha fractions and rearranges them into compounds containing benzene rings. Hydrogen is produced as a by-product of the reactions.
Title: Re: need help with chem AS
Post by: nid404 on February 22, 2010, 07:30:22 pm
in simple words....cyclohexane is not a product of reforming