Solve for the correct answers and check the correct answers Moles of CO₂ = (Mass of CO₂) ÷ (Molar Mass of CO₂)0.781/44.01 = 0.0177Molar Volume = VSTP ÷ Moles of CO₂Percent error = = 1-432/0.0177 = 24.486 L/molexperimental value-true valuetrue valueTrue value = 22.4 L/mol22.4 - 24.40622.4| x 100*100= 8.955% Here's the first problem. You are doing an experiment in a laboratory that is 24 °C and 0.980atm of pressure. You take a 0.480L flask that weighs 172.201g and fill it with CO2. After it isfilled it weighs 172.406g. Calculate the molar volume of the CO₂!Vsip =P₁V₁TSTPT1PSTP0.980x0.480* 273'297 * 1128.4192297= 0.432L1.2000-480 =0.5769mass of 20₂ = 172.406-172-201 + 0.576 =2.0.78190-781/44.01 = 0.0177 mol Co₂le volume 0.432/0.0177 = 24.486 L/molNow, we need to calculate the mass of the air that was inside the flask before we filled it withCO₂:Mass of air = (density of air)(volume of flask)1.200 * 0.480 = 0.5769Now, calculate the mass of the CO₂ by subtracting the mass of the empty flask from the mass ofthe filled flask, adding the mass of air19Mass of CO₂ = (Mass of flask filled with CO₂) - (Mass of empty flask) + (Mass of air)172-406 - 172.201 + 0.576 = 0.7816

I think you have solved some of them correctly but mostly are wrong due to a minor mistake. Here, I…

es of CO₂ = (Mass of CO₂) ÷ (Molar Mass of CO₂) 0.781/44.01 = ar Volume = VSTP + Moles of CO₂ ent error = | 432/0.0177 = 24.406 L/mol experimental value-true value true value value = 22.4 L/mol 0.01779 22.4-24:406 | x 100 *100

es of CO₂ = (Mass of CO₂) ÷ (Molar Mass of CO₂) 0.781/44.01 = ar Volume = VSTP + Moles of CO₂ ent error = | 432/0.0177 = 24.406 L/mol experimental value-true value true value value = 22.4 L/mol 0.01779 22.4-24:406 | x 100 *100

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter10: Solutions

Section: Chapter Questions

Problem 89QAP

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