Saved Bookmarks
| 1. |
`4.0 g` of `NaOH` is contained in one decilitre of aqueous solution. Calculate the following in the solution (d of `NaOH` solution `= 1.038 g mL^(-1)`) a. Mole fraction of `NaOH` b. Molartiy of `NaOH` c. Molality of `NaOH` |
|
Answer» `M = (W_(2) xx 100)/(Mw_(2) xx V_(sol))` (1 decilitre `= 100 mL`) `= (4 xx 1000)/(40 xx 100) = 1` `m = (W_(2) xx 1000)/(Mw_(2) xx ("Weight of solution - Weight of solute")` `W_(1)` = weight of solvent = Weight of solution - weight of solute `= V_(sol) xx d_(sol) - W_(2)` `= 100 xx 1.038 - 4 = 99.8 g` `:. m = (4 xx 1000)/(40 xx 99.8) = 1.002` `X_(2) = (n_(2))/(n_(1) + n_(2)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))` `= (4 // 40)/((99.8)/(18) + (4)/(40)) = (0.1)/(5.54 + 0.1) = 0.0177` |
|