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kenny6666 [7]
3 years ago
8

If the van't Hoff factor for LiCl in a 0.62m solution is 1.92, what is the vapor pressure depression in mmHg of the solution at

298 K? (The vapor pressure of water at 298 K is 23.76 mmHg.)
Chemistry
1 answer:
krek1111 [17]3 years ago
7 0

Answer:

0.50mmHg

Explanation:

Raoult's law (Vapor pressure depression) is:

P_{solution} = X_{solvent} P^0_{solvent}

Where P is vapour pressure of solution and vapour pressure of solvent and X is mole fraction of solvent.

Van't Hoff factor helps to know how many species comes from the dissolution of the compound in the solvent.

Assuming volume of solution is 1L, moles of LiCl in solution are:

0.62×1.92 = 1.19 moles of LiCl in solution

Moles of water in 1L are:

1L ≡ 1000g × (1mol / 18g) = 55.56mol H₂O

Mole fraction of water is:

55.56mol / (55.56mol + 1.19mol) = 0.979

Vapour pressure of solution is:

P_{solution} = 23.26mmHg

That means vapour pressure depression is:

23.76mmHg - 23.26mmHg = <em>0.50mmHg</em>

<em></em>

I hope it helps!

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Answer:

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The balanced equation for the reaction between zinc and acetic acid is

Zn(OH)_2 +2CH_3 COOH ----> Zn(CH_3COO)_2+2H 2O

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A balanced chemical equation is defined as the chemical equation which has an equal number of atoms on both sides of the reaction arrow. A balanced chemical equation is based on the law of conservation of mass.

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Below is the reaction between acetic acid and Zinc:  

Z n  +  2 H C 2 H 3 O 2  →  Z n ( C H 3 C O O ) 2  +  H 2

This reaction is an example of a reaction between an acid and a metal. When an acid reacts with a metal, salt, and Hydrogen gas forms. The Hydrogen gas formed here can be tested as it burns with a POP sound.

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when 6g acetic acid is dissolved in 1000cm3 of solution then how many molecules ionize out of 1000 acetic acid molecules
iVinArrow [24]

Answer:

24.8 molecules are ionized from 1000 acetic acid molecules.

Explanation:

Acetic acid, CH₃COOH dissociates in water, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

<em>That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.</em>

<em />

The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:

6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.

0.1 moles / L = <em>0.1M</em>

The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:

[CH₃COOH] = 0.1M - X

[CH₃COO⁻] = X

[H⁺] = X

Replacing in Ka formula:

6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

6.3x10⁻⁵ = [X] [X] / [0.1 - X]

6.3x10⁻⁶ - 6.3x10⁻⁵X = X²

6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0

Solving for X

X = - 0.0025 → False solution, there is no negative concentrations.

X = 0.00248M

That means, a 0.1M of acetic acid produce:

[CH₃COO⁻] = X = 0.00248M solution of the ionized form.

In a basis of 1000 molecules:

1000 molecules × (0.00248M / 0.1M) = 24.8

<h3>24.8 molecules are ionized from 1000 acetic acid molecules.</h3>
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