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