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what's the concentration of this?

I'm using the formula Absorbance = Ecl and need c in moldm^-3 i think.

i have 5cm^3 methyl orange and added water up to 25cm^3, how do i express this as moldm^-3? do i have enough info to do this?

If not, i have a graph of x / 25cm3 on the x axis ( where x is how much methyl orange in 25cm3 solution), and absorbance on y axis, with a line of best fit. Can i use conc = 5, if x = 5, in formula? (I'm finding E)

Can i use density = mass/vol to find number of molecules then moles?

looks like i should know how to do it but i can't find how to. Thanks.
I would think that that would be enough information,

As you said, you could use density(p)=mass(m)/vol(v). From google the density of methyl orange is 1.28gcm-3,
so m=pv =1.28*5=6.4g

From google again, the GFM of methyl orange is 327.33gmol-1
so the number of moles(n)=m/GFM = 6.4/327.33 = 0.01955...mol in the 5cm3

This becomes 0.01955...mol in 25cm3 when the water is added,
25cm3=0.025dm3 because 1cm3=0.001dm3,
so that's 0.01955...mol in 25cm3, or 0.01955... mol in 0.025dm3,
therefore that's 0.782moldm-3 (0.01955.../0.025)

I think that that should be right :smile: I hope that that helps!!

(This would only work if the 5cm3 were 5cm3 of the solid methyl orange :/ I'm not sure how to do it otherwise without knowing the concentration of the initial solution)
(edited 8 years ago)
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Nat_LPS
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Original post by ffiorentini
I would think that that would be enough information,

As you said, you could use density(p)=mass(m)/vol(v). From google the density of methyl orange is 1.28gcm-3,
so m=pv =1.28*5=6.4g

From google again, the GFM of methyl orange is 327.33gmol-1
so the number of moles(n)=m/GFM = 6.4/327.33 = 0.01955...mol in the 5cm3

This becomes 0.01955...mol in 25cm3 when the water is added,
25cm3=0.025dm3 because 1cm3=0.001dm3,
so that's 0.01955...mol in 25cm3, or 0.01955... mol in 0.025dm3,
therefore that's 0.782moldm-3 (0.01955.../0.025)

I think that that should be right :smile: I hope that that helps!!

(This would only work if the 5cm3 were 5cm3 of the solid methyl orange :/ I'm not sure how to do it otherwise without knowing the concentration of the initial solution)


that makes sense, but i found out it was a solution at the start, and i have no idea its concentration, i guess i've learned to look for one next time aha. thanks a lot though :biggrin:
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Nat_LPS
OP
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Original post by Nat_LPS
that makes sense, but i found out it was a solution at the start, and i have no idea its concentration, i guess i've learned to look for one next time aha. thanks a lot though :biggrin:


Do you know anything about K*_D ? i'm looking at dissociation of a halide ion between 2 layers but i have no idea how to work out one of the concentrations needed

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