Given solutions of $(i)$ $(i)$ $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ ; $(i i)$ $(ii)$ $B a C l_{2}$ $BaCl_2$ ; $(i i i)$ $(iii)$ $N a_{2} S O_{4} (a q)$ $Na_2SO_4(aq)$ ; $(i v)$ $(iv)$ $equal volumes of (ii) and (iii)$ $"equal volumes of (ii) and (iii)"$ ........which preparation should exert the MOST osmotic pressure?

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Given solutions of $(i)$ $(i)$ $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ ; $(i i)$ $(ii)$ $B a C l_{2}$ $BaCl_2$ ; $(i i i)$ $(iii)$ $N a_{2} S O_{4} (a q)$ $Na_2SO_4(aq)$ ; $(i v)$ $(iv)$ $equal volumes of (ii) and (iii)$ $"equal volumes of (ii) and (iii)"$ ........which preparation should exert the MOST osmotic pressure?

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Answer 1 · 2017-06-02T06:32:34

We gots $(i)$ $(i)$ $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ ; $(i i)$ $(ii)$ $B a C l_{2}$ $BaCl_2$ ; $(i i i)$ $(iii)$ $N a_{2} S O_{4} (a q)$ $Na_2SO_4(aq)$ ; $(i v)$ $(iv)$ $equal volumes of (ii) and (iii)$ $"equal volumes of (ii) and (iii)"$ ........I suspect it is solution $(i v)$ $(iv)$

Explanation:

And $(i v)$ $(iv)$ a solution obtained by mixing equal volumes of $(i i)$ $(ii)$ and $(i i i)$ $(iii)$ .

Now the highest osmotic pressure will be expressed by the solution which CONTAINS LEAST number of ions..........

And $(i)$ $(i)$ has got 5 ions in solution, $2 \times A l^{3 +}$ $2xxAl^(3+)$ , and $3 \times S O_{4}^{2 -}$ $3xxSO_4^(2-)$ .................

And $(i i)$ $(ii)$ has got 3 ions in solution, $2 \times C l^{-}$ $2xxCl^(-)$ , and $B a^{2 +}$ $Ba^(2+)$ .................

And $(i i i)$ $(iii)$ has got 3 ions in solution, $2 \times N a^{+}$ $2xxNa^(+)$ , and $S O_{4}^{2 -}$ $SO_4^(2-)$ .................

But $(i v)$ $(iv)$ mixes $(i i)$ $(ii)$ and $(i i i)$ $(iii)$ , and you know, or should know that $B a S O_{4}$ $BaSO_4$ is pretty insoluble stuff. (All sulfates are soluble except for $H g S O_{4}$ $HgSO_4$ , and $C a S O_{4}$ $CaSO_4$ , and $B a S O_{4}$ $BaSO_4$ .

And thus, the following reaction occurs........

$B a^{2 +} + S O_{4}^{2 -} \to B a S O_{4} (s) ⏐ ⏐ ↓$ $Ba^(2+) + SO_4^(2-)rarr BaSO_4(s)darr$

In this solution, there remain 2 equiv each of $N a^{+}$ $Na^+$ and $C l^{-}$ $Cl^-$ . As far as I can tell (and you have not set out the problem as clearly as I would like), because $B a S O_{4}$ $BaSO_4$ precipitates, this solution, $(i v)$ $(iv)$ , has the LEAST number of ions in solution, and thus the highest osmotic pressure.

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Given solutions of (i)(i) Al2(SO4)3Al_2(SO_4)_3; (ii)(ii) BaCl2BaCl_2; (iii)(iii) Na2SO4(aq)Na_2SO_4(aq); (iv)(iv) equal volumes of (ii) and (iii)"equal volumes of (ii) and (iii)"........which preparation should exert the MOST osmotic pressure?

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