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The expression relating mole fraction of solute `(chi_(2))` and molarity `(M)` of the solution is: (where `d` is the density of the solution in `g L^(-1)` and `Mw_(1)` and `Mw_(2)` are the molar masses of solvent and solute, respectivelyA. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) + 1000d)`B. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) + d)`C. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) - 1000d)`D. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) - d)` |
Answer» Correct Answer - B Let the volume of solution `= 1 L` Weight of soulition `= 1 xx d = dg` Number of moles of solute `(n_(2))` in `1 L` solution `= M` `:. (W_(2))/(Mw_(2)) = M` `W_(2)` (weight of solute) `= M xx Mw_(2)` `W_(1)` (weight of solvent) = Weight of solution - Weight of solute `= (d - M xx Mw_(2))` Number of moles of solvent `(n_(1))` `= (W_(1))/(Mw_(1)) = ((d - M xx Mw_(2))/(Mw_(1)))` `chi_(2) = (n_(2))/(n_(1) + n_(2)) = (M)/(((d - M x Mw_(2))/(Mw_(1))) + M)` `= (M xx Mw_(1))/(M (Mw_(1) - Mw_(2)) + d)` |
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