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Vladimir79 [104]
3 years ago
11

At 55.0 ∘C, what is the vapor pressure of a solution prepared by dissolving 78.4 g of LiF in 253 g of water? The vapor pressure

of water at 55.0 ∘C is 118 mmHg. Assume complete dissociation of the solute. Express your answer to three significant figures and include the appropriate units.
Chemistry
1 answer:
ddd [48]3 years ago
7 0

Answer:

82.5 mmHg

Explanation:

The vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent and on the vapor pressure of the pure solvent at the same temperature.

P sol=  χsolvent ⋅ P∘solvent, where

Psol - the vapor pressure of the solution

χ solvent - the mole fraction of the solvent

P ∘solvent - the vapor pressure of the pure solvent

Now, the most important thing to realize here is that lithium fluoride,  

LiF , a soluble ionic compound, will dissociate completely in aqueous solution to form lithium cations and fluoride anions.

LiF (aq] → Li+(aq]  + F − (aq]

Notice that every mole of lithium fluoride produces one mole of lithium cations and one mole of fluoride anions.

This means that you get a total of two moles of ions for every one mole of lithium fluoride in the solution.

Use lithium fluoride's molar mass to determine how many moles would be found in the given sample

78.4g⋅ (1 mole LiF /25.94 g)  = 3.022 moles LiF

This means that the solution will contain

3.022 moles LiF ⋅  (2 moles ions /1 mole LiF)  = 6.044 moles ions

Now, water's mole fraction in this solution will be equal to the number of moles of water divided by the total number of moles present in the solution.

To get the number of moles of water, use its molar mass

253  g  ⋅  (1 mole H 2O / 18.015 g) = 14.044 moles H 2O

This means that water's mole fraction will be

χwater = 14.044 moles  / (14.044 +6.044 )moles = 0.6991

The vapor pressure of the solution will thus be

Psol= 0.6991  ⋅ 118 mmHg = 82.5 mmHg

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The half-life of the reaction is 50 minutes

Data;

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<h3>Reaction Constant</h3>

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x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\

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Answer:

2.79 °C/m

Explanation:

When a nonvolatile solute is dissolved in a pure solvent, the boiling point of the solvent increases. This property is called ebullioscopy. The temperature change (ΔT) can be calculated by:

ΔT = Kb*W*i

Where Kb is the ebullioscopy constant for the solvent, W is the molality and i is the van't Hoff factor.

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Where m1 is the mass of the solute (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).

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W = 2/(147*0.0225)

W = 0.6047 mol/kg

(82.39 - 80.70) = Kb*0.6047*1

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