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

Arun
22545 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))   RegardsArun (askIITians forum expert)
```
one year ago
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