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Anna35 [415]
3 years ago
13

For this exercise, we will consider a solution of the nonvolatile solvent ethylene glycol, C2H4(OH)2, in water. If the solution

contains 1 mol of C2H4(OH)2 for each 4 mol of H2O, what is the mole fraction of water?
Chemistry
1 answer:
Lemur [1.5K]3 years ago
3 0

Answer:

Mole fraction of water is 0.8.

Explanation:

Mole fraction is defined as ratio of moles of single component to the sum of moles of all components present in the solution.

Moles of ethylene glycol = n_1=1 mol

Moles of water = n_2=4 mol

Mole fraction of ethylene glycol = \chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1 mol}{1 mol+4 mol}=0.2

Mole fraction of water  = \chi_2=\frac{n_2}{n_1+n_2}

\chi_2=\frac{4 mol}{1 mol+4 mol}=0.8

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Answer:

A)Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

B)Ksp=1.75×10^−10

C)Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=​Δ​G​f​,​p​r​o​d​u​c​t​s​∘​−​Δ​G​f​,​r​e​a​c​t​a​n​t​s​∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​C​l​(​a​q​)​−​∘​]​−​Δ​G​f​,

A​g​C​l​(​s​)​⇌​A​g​(​a​q​)​+​+​C​l​(​a​q​)​−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​B​r​(​a​q​)​−​∘​]​−​Δ​G​f​,

Δ​G​r​x​n​∘​=​[​77.1​k​J​/​m​o​l​+​(​−​104.0​k​J​/​m​o​l​)​]​−​(​−96.90kj/mol

Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K

Ksp=5.45×10^−13

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3 years ago
Abu Ali al-Hasa invented the words____and___
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In standardizing a naoh solution a student found that 25.55cm of base neutralize exactly 21.35cm of 0.12M HCl find the molarity
nalin [4]

Answer:

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Explanation:

First let's generate a balanced equation for the reaction

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From the equation,

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Mb = Molarity of base =?

We obtained nA(mole of acid) and nB(mole of base) to be 1

The molarity of the base can be calculated for using:

MaVa/ MbVb = nA / nB

0.12x21.35 / Mb x 25.55 = 1

Cross multiply to express in linear form

Mb x 25.55 = 0.12x21.35

Divide both side by 25.55

Mb = (0.12x21.35) / 25.55

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Explanation:

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