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Method: A reflux apparatus was set up consisting of a condenser,pear-shaped flask, thermometer, condenser and hot plate. Into thepear-shape flask, 10.0cm3 of one li...

Question

Method: A reflux apparatus was set up consisting of a condenser,pear-shaped flask, thermometer, condenser and hot plate. Into thepear-shape flask, 10.0cm3 of one liquid was placed andheated until a slow steady reflux is obtained. The boiling pointwas recorded. The liquid was allowed to cool before adding2.0cm3 of the second liquid via measure cylinder. Themixture was heated until a slow steady reflux was obtained. Theboiling point were recorded after further successive additions of2.0cm3 portion

Method: A reflux apparatus was set up consisting of a condenser, pear-shaped flask, thermometer, condenser and hot plate. Into the pear-shape flask, 10.0cm3 of one liquid was placed and heated until a slow steady reflux is obtained. The boiling point was recorded. The liquid was allowed to cool before adding 2.0cm3 of the second liquid via measure cylinder. The mixture was heated until a slow steady reflux was obtained. The boiling point were recorded after further successive additions of 2.0cm3 portions until a total of 10.0cm3 of the second liquid had been added. The experiment was repeated starting with 10.0cm3 of the second liquid in the flask and 2.0cm3 additions of the first liquid. QUESTIONS: A. COMPLETE THE TABLES BELOW: Mixture Boiling Point Total Vapour Pressure (KPa) Volume of Acetone (mL) Volume of Chloroform (mL) Mole Fraction of Acetone Mole Fraction of Chloroform ℃ K 10 0 57.0 101.325 10 2 60.0 101.325 10 4 62.0 101.325 10 6 63.5 101.325 10 8 64.0 101.325 10 10 65.0 101.325 0 10 61.0 101.325 2 10 63.0 101.325 4 10 63.5 101.325 6 10 64.0 101.325 8 10 63.0 101.325 10 10 63.0 101.325 Mixture Boiling Point Total Vapour Pressure (KPa) Volume of pentane (mL) Volume of hexane (mL) Mole Fraction of pentane Mole Fraction of hexane ℃ K 10 0 38.0 101.325 10 2 45.0 101.325 10 4 49.0 101.325 10 6 52.0 101.325 10 8 54.5 101.325 10 10 56.0 101.325 0 10 68.0 101.325 2 10 63.0 101.325 4 10 59.5 101.325 6 10 56.5 101.325 8 10 55.0 101.325 10 10 55.0 101.325 B. Show the calculation used to obtain the mole fraction of acetone and chloroform when 10ml acetone and 6ml chloroformed are mixed. (density of acetone = 0.79 g/cm3; density of chloroform = 1.48g/cm3) C. Plot a graph of boiling Points (℃) against composition by mole fraction for a binary mixture of acetone and chloroform. D. Plot a graph of boiling Points (℃) against composition by mole fraction for a binary mixture of pentane and hexane. E. Based on the graph of boiling points against mole fraction by mole fraction for a binary mixture of acetone and chloroform, does this binary mixture obeys Raoult’s Law? Justify your answer. What does this mean for the intermolecular forces of the binary mixture? F. Can a mixture of liquids be purified using simple distillation? Justify your position.



Answers

The feed to a distillation column (sketched below) is a 45.0 mole\% $n$ -pentane- 55.0 mole\% n-hexane liquid mixture. The vapor stream leaving the top of the column, which contains 98.0 mole\% pentane and the balance hexane, goes to a total condenser (which means all the vapor is condensed). Half of the liquid condensate is returned to the top of the column as reflux and the rest is withdrawn as overhead product (distillate) at a rate of $85.0 \mathrm{kmol} / \mathrm{h}$. The distillate contains $95.0 \%$ of the pentane fed to the column. The liquid stream leaving the bottom of the column goes to a reboiler. Part of the stream is vaporized; the vapor is returned to the bottom of the column as boilup, and the residual liquid is withdrawn as bottoms product.(a) Calculate the molar flow rate of the feed stream and the molar flow rate and composition of the bottoms product stream. (b) Estimate the temperature of the vapor entering the condenser, assuming that it is saturated (at its dew point) at an absolute pressure of 1 atm and that Raoult's law applies to both pentane and hexane. Then estimate the volumetric flow rates of the vapor stream leaving the column and of the liquid distillate product. State any assumptions you make. (c) Estimate the temperature of the reboiler and the composition of the vapor boilup, again assuming operation at 1 atm.(d) Calculate the minimum diameter of the pipe connecting the column and the condenser if the maximum allowable vapor velocity in the pipe is $10 \mathrm{m} / \mathrm{s}$. Then list all the assumptions underlying the calculation of that number.

So just drop of some contextual information. I'll just draw up on the screen. The equations of interest to have log he started the bay for pressure is equal to a minus. B over T f C ABC The coefficients t is all temperature we have pieced are a since Y a multiplied by P. P start is a partial pressure. Why is the mole fraction of Pete is the total pressure? So the full form of L. F Ellis, the lower flammability limit. So what we have is a flashpoint of us over a 15 flashpoint Brussels based 75 degrees Celsius and in the ambient temperature of our lab is 20 to 25 degrees Celsius. So the temperature of solving a is 15 degrees Celsius, which is lower than the ambient temperature in the lab so soften a can ignite easily have exposed exposed to a flame. However, if we take a look at solving B, it's 75 degrees Celsius, which is higher than the ambient temperature in the lab. To solve and be cannot ignite is easily if exposed to a flame, so solvent be is safer to keep in the lab two. Therefore, it must be monitored the temperature of the solvent. They does not rise above 15 degrees Celsius streets, which it must be refrigerated. However, there is no need to put solvent be in the fridge. It's moving on, Susie next part here. So the LF l of methanol in air is about six miles to sound, and the total pressure is one atmosphere. So one atmosphere we know is equal to 100 and 1.3 to 5. Killer Paschal's more fraction of methanol is equals the ratio one component in the total mixture. So why, for methanol is equal to 9.6 fried by 100 we get not quite, not six to the mole. Fraction of math does not point North. Six. We can calculate the partial pressure of methanol. You start. Um, it's equal to not point not six. We'll swap by one a 1.3 to 5. That gives us 6.795 kill of our skulls. So the part of pressure methane is 6.0 76 95 Killer pass girls. I was just written on the screen there. So now that we know the partial pressure of methanol, methanol coefficients. We have listed as a B and C weaken. Substitute them in on solve our equation. Hands we have logged 6.795 seek to 16.5785 Take away 3638.27 Divided by T. R. 239.5 We sold the T. We got a value of 6.768 degrees Celsius, so the flashpoint of methanol and the temperature are closer to each other. Therefore, the values of flashpoint were taken around 7 to 8 degrees Celsius. Eso, the flashpoint of methanol, is a lot less than the ambient temperature of the environment. So if methanol is exposed to an open flame, it will easily evaporate. Entertain? LF l and if it's exposed to ignition source, it can easily catch fire. So the common ignition sources we have our sparks heart plates, Bunsen burners in the lab

So in the first part of this podcast, Wigan, using the T X Y diagrams, which is temperature of us, is more fraction in liquid and vapor pressure phase for a component that is often an easier way to calculate the bullet points the two points in the relative compositions. So we need to look at the T X Y diagram for benzene in the first part here. So we find that the first drop of vapor condenses at 104 degree C and then we move horizontally. From this point to the liquid carbon, determine the mole. Fraction of benzene in the liquid is not 0.0.1. For on that, the more fraction of tolerant and liquid is no point 86 So in the second part here we have the mole fraction of benzene in the liquid and vapor 100 degrees, so we can use the basis of 1.0 mile of original paper, and then we can write equations for the mall balance of fanzine. So what we have is why be not multiplied by NT? People too wide be multiplied by and V x B was by by an Al. So we have not 0.30 multiplied by 1.0 miles, the equal to not 0.46 and be at no point to for an elf. So I will jump onto a fresh page here. So in Vienna on LR numbers of moles of vapor liquid and why the note is a more fraction of benzene in the original vapor. So the total more violence we have NT is equal to M V at N. L. So NT is essentially called one more yourself envy on an L. So we know that an l table 2.73 more and B is not point to seven more. No 0.27 divided by 9.73 is it's not 0.37 when we divide them by one another. So for this last part here, we need to take a look at the T X Y diagram again for benzene. We moved down vertically from the full vapor, with a 9.3 more fraction of benzene at 115 degrees Celsius. That's our temperature now on the line, a intersects with the liquid curve at the temperature of 98 degrees Celsius temperature, so that is the last drop of vapor that condenses at 98 degrees Celsius. So then we move horizontally from that point to the vapor curve, and we can see that the mole fraction of fanzine in the paper is not 0.53 My


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