Calculate the molality `(m)` of `3 M` solution of `NaCl` whose density is `1.25 g mL^(-1)`.

Explanation

Molar mass `(Mw_(2))` of `NaCl = (23 + 35.5) g = 58.5 g`
`M = 3 mol L^(-1)`
Mass of weight `(W_(2))` of `NaCl` in `1 L` solution
`3 xx 58.5 = 175.5 g`
First method:
`d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))`
`1.25 g mL^(-1) = 3 ((58.5)/(1000) + (1)/(m))`
`(1.25)/(3) (58.5)/(1000) = (1)/(m)`
`0.416 - 0.058 = (1)/(m)`
`0.358 = (1)/(m)`
`m = (1)/(0.358) = 2.79 m`
second method:
`M = 3 mol L^(-1)`
Mass of `NaCl` in `1 L` solution `(W_(2)) = 3 xx 5835 = 175.5 g`
Mass of `1 L` solution `= V_(sol) xx d_(sol)`
`= 1000 xx 1.25 = 1250 g`
Mass of `H_(2) O` in solution `(W_(1))`
Mass of solution - Mass of slute
`= W_(sol) - W_(2)`
`= 1.250 - 175.5 = 1074 g = 1.0754 kg`
`:. m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))`
`= (175.5 xx 1000)/(58.5 xx 1074.5) = 2.79 m`
or `m = ("Number of moles of solute")/("Mass of solvent in kg")`
`= (3 mol)/(1.0745 kg) = 2.79 m`