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barxatty [35]
2 years ago
10

Na3po4 dissolves in water to produce an electrolytic solution. what is the osmolarity of a 2. 0 × 10-3 m na3po4 solution?

Chemistry
1 answer:
vichka [17]2 years ago
6 0

When Na3po4 dissolves in water to produce an electrolytic solution. The osmolarity of a 2. 0 × 10-3 m Na3po4 solution is 0.008osmol/L.

Osmolarity is defined as the number of osmoles of solute particles per unit volume of the solution.

In other words osmolarity is the multiple if molarity

Osmolarity = i× molarity

Here i represents the van't Hoff factor,

Na_{3}PO_{4} ⇒ 3Na^{+} + PO_{4} ^{-}

3  Moles of Na_{} ^{+} + 1 mole PO_{4} ^{-} = 4

The number of moles of particles of solute produced in solution are actually called osmoles.

As a result, the van't Hoff factor will be equal to

i=4 Moles ions produced (osmoles) 1mole Na_{3} PO_{4} .dissolved =4

Since we know that,

Na_{3} PO_{4} = 2.0 * 10^{-3} M

Osmolarity =

4*2.0*10^{-3} M = 8.0 * 10^{-3}  osmol L -10s

Thus, the Osmolarity of given solution is 0.008 osmol/L.

learn more about Osmolarity:

brainly.com/question/13597129

#SPJ4

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Answer :The corrected answer is given below;


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Hydrochloric acid and sulfuric acid are <u>completely</u> ionized in solution and are <u>strong</u> acids. Ethanoic acid, which is only about 1 percent ionized, is a <u>weak</u> acid. Magnesium hydroxide and calcium hydroxide are strong <u>base</u>.

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3. If x electrons are needed to displace 108 g silver from a solution which contains Ag ions,
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Answer:

Choice A. x electrons would be required for displacing 9\; \rm g of aluminum from a solution of \rm Al^{3+} ions.

Assumption: by "\rm Ag ions" the question meant \rm Ag^{+} with a charge of +1 on each ion.

Explanation:

The question states that the relative atomic mass of \rm Ag is 108. In other words, each mole of

Therefore, that 108\; \rm g\! of silver that were formed would contain 1\; \rm mol of silver atoms.

Metallic silver would precipitate out of this \rm Ag^{+} solution only after these ions are turned into \rm Ag atoms.

One \rm Ag^{+} ion carries one unit of positive electrical charge. On the other hand, each  e^{-} carries one unit of negative electrical charge.

Therefore, each \rm Ag^{+}\! ion will need to gain one electron to form a neutral \rm Ag atom.

{\rm Ag^{+}}\; (aq) + e^{-} \to {\rm Ag}\; (s).

At least  1\; \rm mol of electrons would be required to turn 1\; \rm mol\! of \rm Ag^{+} ions into that 1\; \rm mol\!\! of silver atoms (which have a mass of 108\; \rm g\!.)

Hence, x = 1\; \rm mol.

Unlike \rm Ag^{+} ions, each aluminum ion \rm Al^{3+} carries three units of positive electrical charge. That is three times the amount of charge on one \rm Ag^{+}\! ion. Therefore, three electrons will be required to turn one \rm Al^{3+}\! ion to an \rm Al atom.

{\rm Al^{3+}}\; (aq) + 3\, e^{-} \to {\rm Al}\; (s)

The question states that the relative atomic mass of \rm Al is 27. Therefore, each mole of \rm Al\! atoms would have a mass 27\; \rm g. There would be \displaystyle \frac{9\; \rm g}{27\; \rm g \cdot mol^{-1}} = \frac{1}{3} \; \rm mol of atoms in that 9\; \rm g of \rm Al\!\!.

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That corresponds to the first choice, x electrons.

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