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Aqueous solution of NaOH is marked 10% `(w//w)`. The density of the solution is 1.070 g `cm^(-3)`. Calculate (i) molarity and (ii) molality of NaOH. |
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Answer» Correct Answer - (i) Molarity of NaOH solution =2. 675 M (ii) Molality of NaOH solution = 2.778 m Given : Percentage by weight of NaOH `=10% (w//w)` Density of NaOH solution =D =1.070 g `cm^(-3)` Molar mass of NaOH =40 g `mol^(-1)` (i) Molarity of NaOH solution = ? (ii)Molality of NaOH solution = ? (i) For molarity of solution : Consider 100 g NaOH solution `:. ` Weight `H_(2)O+` Weight of NaOH =100 g `:. ` Weight of `H_(2)O = 100 - 10 = 90 g` Number of moles of NaOH `= (W_(NaOH))/(M_(NaOH))` `:. n_(NaOH) = .(10)/(40)` `=0.25 `mol Density of solution `(D) = ("Weight of solution " )/(" volume of solution " ) = (W)/(V)` `:. V = (W)/(D)` `=(100)/(1.070)` `=93 . 46 cm^(3)` `=0.09346 dm^(3)` Molarity `=(n)/(V)= (0.25)/(0.09346) =2.675 "mol" dm^(-3) (or M)` (ii) For molality of NaOH solution : Molality `=("Molar of NaOH ") /("Mass of solvent in gram ") xx 1000` `=(0.25)/(90) xx 1000` `=2.779 "mol" kg^(-1) (or m)` |
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