Question

The solubility of $Ba{\left(OH\right)}_2$. $8H_{2}O$ in water ar 288K is 5.6g per 100g of water. What is the molality of the hydroxide ions in saturated solution of $Ba{\left(OH\right)}_2$. $8H_{2}O$ at 288K?

[ At. mass of Ba = 137, O = 16, H = 1]

• A. 0.869
• B. 1.32
• C. 0.355
• D. None of these

Answer 1

The correct option is C 0.355

The molecular mass of $Ba(OH{)}_2=315.4639g/mol$

It is given that 5.6g in 100 g of water which is same as 56g in 1 kg of water.

So, number of moles of $Ba(OH{)}_2$ in 1 kg of water $=\frac{56}{315.4639}=0.1775mol$.

The dissociation of $Ba(OH{)}_2$ is,

$Ba(OH{)}_{2}8.H_{2}O\to B{a}^{2+}+2O{H}^{-}+8H_{2}O$

1 mole of $Ba(OH{)}_2$ produces 2 moles of $O{H}^{-}$ions

So the number of moles of $O{H}^{-}$ ions $=2\times 0.1775=0.355$ moles.

Therefore the molality of $O{H}^{-}$ ions $=0.355m$.