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Shown below are the results of water quality analyses of the Mississippi River at Baton Rouge, Louisiana. If the water is treated with 30.00 mg/L of ferric chloride for turbidity coagulation, how much alkalinity will remain? Ignore the side reactions with phosphorus and assume all the alkalinity is HCO3\mathrm{HCO}_{3}^{-}. Mississippi River, Baton Rouge, Louisiana

ConstituentExpressed asMilligrams per literTotal hardness CaCO3164.0Calcium hardness CaCO3108.0Magnesium hardness CaCO356.0Total ironFe0.9CopperCu0.01ChromiumCr0.03Total alkalinity CaCO3136.0ChlorideCl32.0Phosphate ( total) PO43.0Silica SiO210.0Suspended solids 29.9 Turbiditya NTUb12.0 pHa 7.6\begin{matrix} \text{Constituent} & \text{Expressed as} & \text{Milligrams per liter}\\ \text{Total hardness} & \ {CaCO_3} & \text{164.0}\\ \text{Calcium hardness} & \ {CaCO_3} & \text{108.0}\\ \text{Magnesium hardness} & \ {CaCO_3} & \text{56.0}\\ \text{Total iron} & \text{Fe} & \text{0.9}\\ \text{Copper} & \text{Cu} & \text{0.01}\\ \text{Chromium} & \text{Cr} & \text{0.03}\\ \text{Total alkalinity} & \ {CaCO_3} & \text{136.0}\\ \text{Chloride} & \text{Cl} & \text{32.0}\\ \text{Phosphate ( total)} & \ {PO_4} & \text{3.0}\\ \text{Silica} & \ {SiO_2} & \text{10.0}\\ \text{Suspended solids} & \text{ } & \text{29.9}\\ \ {Turbidity^a} & \ {NTU^b} & \text{12.0}\\ \ {pH^a} & \text{ } & \text{7.6}\\ \end{matrix}

aNot^aNot in mg/L. bNTU^bNTU = nephelometric turbidity units and the pH is dimensionless.

Question

Give an answer for the follwoing exercise.

(a) Trouton's rule states that the ratio of the molar heat of vaporization of a liquid (ΔHvap )\left(\Delta H_{\text {vap }}\right) to its boiling point in kelvins is approximately 90 J/Kmol90 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}. Use the following data to show that this is the case and explain why Trouton's rule holds true.

Tbp(C)ΔHvap (kJ/mol) Benzene 80.131.0 Hexane 68.730.8 Mercury 35759.0 Toluene 110.635.2\begin{array}{lcc} & \boldsymbol{T}_{\mathbf{b p}}\left({ }^{\circ} \mathbf{C}\right) & \boldsymbol{\Delta} \boldsymbol{H}_{\text {vap }}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) \\ \hline \text { Benzene } & 80.1 & 31.0 \\ \text { Hexane } & 68.7 & 30.8 \\ \text { Mercury } & 357 & 59.0 \\ \text { Toluene } & 110.6 & 35.2 \end{array}

(b) Use the values in the table to calculate the same ratio for ethanol and water. Explain why Trouton's rule does not apply to these two substances as well as it does to other liquids.

 Substance  Boiling point (C)ΔHvap (kJ/mol) Argon (Ar)1866.3 Benzene (C6H6)80.131.0 Ethanol (C2H5OH)78.339.3 Diethyl ether (C2H5OC2H5)34.626.0 Mercury (Hg2)35759.0 Methane (CH4)1649.2 Water (H2O)10040.79\begin{array}{ccc} \text { Substance } & \text { Boiling point }\left({ }^{\circ} \mathrm{C}\right) & \Delta H_{\text {vap }}(\mathrm{kJ} / \mathrm{mol}) \\ \text { Argon }(\mathrm{Ar}) & -186 & 6.3 \\ \text { Benzene }\left(\mathrm{C}_6 \mathrm{H}_6\right) & 80.1 & 31.0 \\ \text { Ethanol }\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) & 78.3 & 39.3 \\ \text { Diethyl ether }\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5\right) & 34.6 & 26.0 \\ \text { Mercury }\left(\mathrm{Hg}^2\right) & 357 & 59.0 \\ \text { Methane }\left(\mathrm{CH}_4\right) & -164 & 9.2 \\ \text { Water }\left(\mathrm{H}_2 \mathrm{O}\right) & 100 & 40.79 \end{array}

Explain why the ratio is considerably smaller than 90 J/Kmol90 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} for liquid HF\mathrm{HF}.

Solution

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In this task, it should be explained why the ratio in Trouton's rule for HF\text{HF} is smaller than most other liquids.

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