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calculate activation energy calculate activation energy calculate activation energy Page LayoutRevlew view Normal No Spacing styles -paragraph Font CHEM 1120 Quiz 1 9/12/18 1. (5 points) Nitrogen, N2, has a solubility of 1.75 x 103g/100 mL H2O at 1.00 atm. What is the solubility of nitrogen in water from air at 4.79 atm? Answer:0.781 x 4.79 x 1.75 x 10A-3 g/100 ml 6.55 x 104-3 2. (10 points) Automotive antifreeze consists of ethylene glycol (C2H02), a nonvolatile nonelectrolyte. Calculate A) the boiling point and B) the freezing p of a 25% w/w solution of ethylene glycol (solute) in H2O (solvent) MWethyileose glycol- 62.1 g/mol, K 1.86 C/m,K 0.51 C/m. Gener for colligative properties, AT -Ksm 3. (10 points) Definitions, Match the following and put in correct order, top to bottom. Example: "3. a, b. e. c Description i.Salting the roads AnswerConcept a. Supersaturated English (US

Explanation / Answer

Solution :-

Q1) Air contains 78.1 % of nitrogen

Therefore fraction of nitrogen in the air = 78.1 % / 100 % = 0.781

Solubility of the nitrogen from air = fraction of N2 * pressure of air * solubility of nitrogen at 1 atm     

                                                              = 0.781 * 4.79 atm * 1.75*10^-3 g/100ml

                                                              = 6.55*10^-3 g/100 ml

Therefore solubility of the nitrogen gas from air is 6.55*10^-3 g per 100 ml water.

Q2)Ethylene glycol mass percent = 25 %

Therefore if we assume 100 g mixture then mass of ethylene glycol = 25%*100g/100%= 25 g

Mass of water = 100g – 25 g = 75 g = 0.075 kg

Lets calculate the moles of ethylene glycol

Moles = mass / molar mass

           = 25 g / 62.1 g per mol

          = 0.403 mol

Molality = moles of solute / kg solvent

               = 0.403 mol / 0.075 kg

              = 5.37 m

Now using the molality we can find the boiling point and freezing point of mixture

Calculating the change in boiling point

Delta Tb= Kb* m

               =0.51 oC/m *5.37 m

          = 2.74 oC

Boiling point of solution = boiling point of pure solvent + delta Tb

                                            = 100 oC + 2.74 oC

                                            = 102.7 oC

Calculating the freezing point of solution

deltaTf= Kf* m

            = 1.86 oC/m * 5.37 m

            = 9.99 oC

Freezing point of solution = freezing point of pure solvent – delta Tf

                                               = 0 oC – 9.99 oC

                                               = -9.99 oC

Therefore freezing point of mixture is -9.99 oC which we can round to -10.0 oC

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