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Anarel [89]
3 years ago
14

the vapor pressure of a naqueous solution is found to be 24.9 mmgh at 25C. what is the mole fraction of solute in this solution?

Chemistry
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

Mole fraction of solute is 0.0462

Explanation:

To solve this we use the colligative property of lowering vapor pressure.

First of all, we search for vapor pressure of pure water at 25°C  = 23.8 Torr

Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.

Formula for lowering vapor pressure is:

ΔP = P° . Xm

Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)

Xm = mole fraction

24.9 mmHg - 23.8 mmHg = 23mmHg . Xm

Xm = (24.9 mmHg - 23.8 mmHg) /  23mmHg

Xm = 0.0462

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Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

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Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

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Thus, from the balanced equation above,

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A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

Therefore, the concentration of Na^+ in the solution is 6.75 M

B. Determination of the concentration of PO4^3- in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

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