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, respectively

Options

A. `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

B

Explanation

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)`