Asked by Riana

1) what mass of lead(ii) acetae is required to be dissolved in a 500.0ml volumetric so that the acetate is equal to 1.48?
Answered by DrBob222
1.48 what?
Answered by Riana
mol/l
Answered by Riana
can you tell me the solution?
Answered by DrBob222
You want how many mols? That's M x L = mols acetate = 1.48 M x 0.500 L = ?
Convert mols acetate to mols Pb(C2H3O2)2.

That is mols acetate x (1 mol Pb(C2H3O2/2 mols C2H3O2) = x
Then g Pb(C2H3O2)2 = x mols x molar mass Pb(C2H3O2)2.
Answered by Riana
i got 197g is that correct. am not sure. is that correct.
Answered by Riana
that part confused me....that is mole acetate x (1 mol Pb(C2H3O2/2 mols C2H3O2) =x
do u mean to divie 1 mole of Pb...by c2h3o2 mol?
Answered by DrBob222
197 won't do it. You can check it. Please do.
mols Pb(Ac)2 = 197/325.3 = 0.605.
mols Ac = 2 x 0.605 = 1.21
Then M = mols/L = 1.21/0.5L = 2.42 M which isn't 1.48M

Answered by DrBob222
No. You multiply mols Ac by 1/2
Answered by Riana
is is 481.4g ?
Answered by Riana
i got it...its 120.3g.........right?
Answered by DrBob222
Right. 120 g is it.
Answered by Riana
hey i have posted another question can you see if you can help me in that?
There are no AI answers yet. The ability to request AI answers is coming soon!