Question

The solubility of Ba(OH)2.8H2O in water at 288 K is 5.6g per 100g of water .What is the molality of the hydroxide ions in saturated solution of Ba(OH)2.8H2O at 298K?? ( atomic mass of Ba=137 , O= 16 , H=1)

Answer 1

the molecular mass of Ba(OH)2= 315.4639 g/mol
Given,
     5.6 g in 100g of water, which is same as 56g in 1 kg of water
Therefore,
     Number of moles of Ba(OH)2 in 1 kg of water = 56/ 315.4639
                                                                             =  0.1775 moles
The dissociation of Ba(OH)2 is as follows:

Answer 2

Answer : The molality of the hydroxide ion in saturated solution is, 0.354m

Solution : Given,

Molar mass $Ba(OH)_2.8H_2O$ = 315 g/mole

As, we are given that 5.6 g per 100 g of water that means 5.6 g of $Ba(OH)_2.8H_2O$ present in 100 g of water

Or we can say that,

56 g of $Ba(OH)_2.8H_2O$ present in 1 Kg of water

Now we have to calculate the moles of $Ba(OH)_2.8H_2O$

$\text{Moles of }Ba(OH)_2.8H_2O=\frac{56}{315}=0.177mole$

That means 0.177 mole of $Ba(OH)_2.8H_2O$ present in 1 Kg of water

The dissociation of $Ba(OH)_2.8H_2O$ will be,

$Ba(OH)_2.8H_2O(aq)\rightleftharpoons Ba^{2+}(aq)+2OH^-(aq)+8H_2O(l)$

From this we conclude that each mole $Ba(OH)_2.8H_2O$ produces 2 mole of $OH^-$ ions

The moles of  $OH^-$ ions = $2\times 0.177=0.354mole$

Molality of  $OH^-$ ions = 0.354 mole per Kg = 0.354 mole/Kg = 0.354m

Therefore, the molality of the hydroxide ion in saturated solution is, 0.354m