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The solubility of `ba(OH)_(2)8H_(2)O` in water at 298 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in a saturatd solution of barium hydroxide at 298 K ? (Atomic mass of Ba =137, O=16) |
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Answer» `"Molality of solution(m)"=("Mass of solute"//"Molarmass of solute")/("Mass of solvent in kg")` Mass of solute, `Ba(HO)_(2)8H_(2)O=5.6g` Molar mass of solute, `Ba(OH)_(2)8H_(2)O=137+2(16+1)+8xx18=315 g mol^(-1)` Mass of solvent (water)= 100g=0.1 kg `Molality (m)=((5.6g)//(315gmol^(-1)))/((0.1kg))=0.18 molkg^(-1)=0.18m` Calculation of molality of the OH ions Barium hydroxide ionies inwater as follows: `Ba(OH)_(2) overset((aq)) Leftrightarrow Ba^(2)+(aq)+2OH(aq)` This shows that the molality of OH ions is twice the molality of `Ba(OH)_(2)=2xx0.18m=0.36m` |
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