GRAVIMETRIC SAC QUESTION HELP!?

[vitaminC]

some last minute study for gravimetric sac tomorrow

a toilet cleaner has NaOCl as an active ingredient. 100ml of cleaner has 7.45mg of NaOCl. In a 750ml sample, what is the number of mole of the Cl atoms?


so far all i have is

no of mole(NaOCl)= 7450g/74= 100.67 mol

i'm not sure what to do next/ eqn to use.

thanks

[Water]

This is what I would do

find the n(NaOCl) = 7.45mg (convert the mg into grams) / M(NaOCl



To convert it into 750ml.

You find the concentration first by dividing it by 0.100L

then you would find the number of mole in 750 sample by multiplying it by .750L




so now you have the new n(NaOCL)


By looking at the compound you know that for every 1 mole of NaOCL, there will be one mole of Cl


Therefore,


n(NaOCL) = n(Cl)

[fady_22]

First, I would find out the mass of NaOCl in 750 ml. To do this, use cross multiplication.
So:
100ml=7.45mg
750ml=x mg
...
x=55.88 mg
Then, find the number of mole of Cl in the 750 ml sample (which is simple as you are given mass, but remember the units is milligrams):
n(Cl)=n(NaOCl)=m/mr=(55.88*10^-3)/(74.5)= 7.5*10^-4 mol

[huaxiadragon]

FIRST note the units. It's in milli grams (assuming you've remembered it right)
therefore its a thousandth of a gram

n(NaOCl)=(7.45 times 10^-3) / (23+16+35.5)=10^(-4) mol

Then, you can reason that since 100ml has this much of NaOCl, then 750ml will have 750/100=7.5 times the amount of NaOCl

It's clear that the mole ratio of NaOCl:Cl= 1:1

Therefore the No. of Moles of Cl
n(Cl)=7.5 times 10^-4
       =7.50 * 10^-4 mol (3 sig figs)

I'm not sure I'm allowed to answer people's SAC question. SO...I won't do it next time if its not allowed.

BTW you need to revise Year 11 Mole concepts, the questions is actually a Yer 11 one, so you need to have some common sense and think it through logically

[huaxiadragon]

Oh man I only went for a drink and I'm 2 post late! DAM.

Dam Water and Fady you are fast

[vitaminC]

FIRST note the units. It's in milli grams (assuming you've remembered it right)
therefore its a thousandth of a gram

n(NaOCl)=(7.45 times 10^-3) / (23+16+35.5)=10^(-4) mol

Then, you can reason that since 100ml has this much of NaOCl, then 750ml will have 750/100=7.5 times the amount of NaOCl

It's clear that the mole ratio of NaOCl:Cl= 1:1

Therefore the No. of Moles of Cl
n(Cl)=7.5 times 10^-4
       =7.50 * 10^-4 mol (3 sig figs)

I'm not sure I'm allowed to answer people's SAC question. SO...I won't do it next time if its not allowed.

BTW you need to revise Year 11 Mole concepts, the questions is actually a Yer 11 one, so you need to have some common sense and think it through logically

how did i manage to get a sac question?
its merely harmless revision

cheers all

[huaxiadragon]

FIRST note the units. It's in milli grams (assuming you've remembered it right)
therefore its a thousandth of a gram

n(NaOCl)=(7.45 times 10^-3) / (23+16+35.5)=10^(-4) mol

Then, you can reason that since 100ml has this much of NaOCl, then 750ml will have 750/100=7.5 times the amount of NaOCl

It's clear that the mole ratio of NaOCl:Cl= 1:1

Therefore the No. of Moles of Cl
n(Cl)=7.5 times 10^-4
       =7.50 * 10^-4 mol (3 sig figs)

I'm not sure I'm allowed to answer people's SAC question. SO...I won't do it next time if its not allowed.

BTW you need to revise Year 11 Mole concepts, the questions is actually a Yer 11 one, so you need to have some common sense and think it through logically

how did i manage to get a sac question?
its merely harmless revision

cheers all

Sorry man, I was half asleep. Uhh....too much Fallout 3 lol