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Grade: 11
        Relation b/w molefraction of solute and molarity of the solution?
10 months ago

Answers : (1)

Arun
14889 Points
							
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)
10 months ago
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