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Relation b/w molefraction of solute and molarity of the solution?
one year ago

Dear student

molarity = moles solute / L solution
mole fraction solute = moles solute / (moles solute + moles solvent)

*******
so if we let
n1 = moles solute
n2 = moles solvent
m1 = mass solute
m2 = mass solvent
mw1 = molecular weight solute
mw2 = molecular weight solvent
ρ = density solution...
V = volume of solution
M = molarity
χ1 = mole fraction solute
χ2 = mole fraction solvent

then..
(1) M = n1 / V
(2) χ1 = n1 / (n1 + n2)

rearranging (1) and substituting into (2)
(3) χ1 = MV / (MV + n2)

since...
(4) ρ = (m1 + m2) / V

and since
(5) n = m / mw
(6) m1 = n1 x mw1
(7) m2 = n2 x mw2

substituting (6) and (7) into (4)
(8) ρ = (n1 x mw1 + n2 x mw2) / V

rearranging (8)
(9) ρ = (n1 x mw1)/V + (n2 x mw2) / V

since n1/V = M
(10) ρ = M x mw1 + (n2/V) x mw2

rearranging (10)
n2 = V x (ρ - M x mw1) / mw2

substituting into (3)
χ1 = MV / (MV + V x (ρ - M x mw1) / mw2)

canceling V..
χ1 = M / (M + (ρ - M x mw1) / mw2)

If you wish to rearrange a bit more
χ1 = (M x mw2) / ((M x mw2) + (ρ - M x mw1))

and of course
χ2 = 1 - χ1 = 1 - (M x mw2) / ((M x mw2) + (ρ - M x mw1))

Regards
Arun (askIITians forum expert)
one year ago
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