chemistry 152

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Hey there, this is 10 questions of chemistry.

CHEM 152 Homework 2 Physical Properties of Solutions 1. A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution. 2. The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent and molality, of the sulfuric acid. 3. Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appropriate solvent for the following substances: grease (C20H42) and potassium iodide (KI). 4. A certain soft drink is bottled so that a bottle at 25°C contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is 4.0 × 10−4 atm, calculate the equilibrium concentrations of CO2 in the soda both before and after the bottle is opened. (The Henry’s law constant for CO2 in aqueous solution is 0.031 mol/L • atm at 25°C.) 5. Calculate the expected vapor pressure at 25°C for a solution prepared by dissolving 158.0 g of common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 cm3 of water. (At 25°C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.) 6. Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g water at 25°C. (The vapor pressure of pure water at 25°C is 23.76 torr.) 7. A solution is prepared by mixing 5.81 g of Potassium Iodide and 100g of H2O at 25°C. What is the vapor pressure of the solution? 8. A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34°C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. (Show your calculation) 9. What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car’s radiator that freezes at Ѹ10.0°F (Ѹ23.3°C)? Assume the density of water is exactly 1.00 g/mL. 10. A chemist is trying to identify a human hormone, which controls metabolism, by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be 0.240°C. Calculate the molar mass of the hormone.

Tutor Answer

KathyBlaq
School: Cornell University

Name Professor Class Date Chemistry Hexane is a non-polar solvent. This is because it has C-H bonds that make it suitable for the grease while methanol is a polar solvent due to its O-H group that makes it suitable solvent for the potassium iodide.PCO2 = kCO2CCO2Psoln = XH2OPoH2OPsoln = XH2OPoH2O P total = XA PoA + XB PoBΔTb = Kb mSoluteΔTf = Kf msoluteΔTf = Kf msolute = 0.0217 = 0.215M = 0.99% = 0.1 = =0.217m 1.230g/ml = 1230 g/l 98.1 * 3.75 = 368g of sulfuric acid 1230g of solution – 368g of sulfuric acid = 862g of water = = 29.91% of H2SO4 = 3.75mol/ 0.863kg H2O = 4.35m PCO2 = 5.0 atm in the open bottle CCO2 = PCO2/ kCO2 = (5.0 atm)/ (32L.atm/mol) = 0.16 mol/L CO2 in the soda reaches equilibrium and the atmospheric CO2, as a result PCO2 = 4.0 x 10-4 atm CCO2 = PCO2 / kCO2 = (4.0 x 10-4 atm) / (32 L.atm/mol) = 1.2 x 10-5 mol/L Determining number of moles first: Nsugar = 158.0 g/342.3 g mol-1 = 0.4616 mol sucrose Density = mass g/Volume cm3 mH2O = dH2O x VH2O = 0.9971 g/cm3 x 643.5 cm3 = 641.6 g nH2O = 641.6 g/18.01 g mol-1 = 35.63 mol H2O XH2O = nH2O/nsucrose + nH2O = 35.63/ 0.4616 + 35.63 = 0.9873 Psoln = XH2OPoH2O = (0.9873) (23.76 torr) = 23.46 torr nH2O = 175.0 g/18.01 g mol-1 = 9.72 mol H2O n Na2SO4 = 35.0 g/142 g mol-1 = 0.246 mol Na2SO4 But Na2SO4 ® 2Na+ + SO-24 Hence, N solute = 3 x 0.246 = 0.738 mol X H2O = nH2O/Nsolute + nH2O = 9.72/ 0.738 + 9.72 = 0.929 P Soln = X H2OPoH2O = (0.929) (23.76 torr) = 22.10 torr nAcetone = 5.81 g/(58.1 g/mol) = 0.100 mol Acetone nChloroform = 11.9 g/(119 g/mol) = 0.100 mol Chloroform XAcetone = XChloroform = 0.500 Hence, P total = (0.500) (345 torr) + (0.500) (293 torr) = 319 torr = (100.34 -100.00) oC = 0.34 oC msolute = ΔTb/ Kb = (0.34 oC)/(0.51 oC. kg/mol) = 0.67 mol/kg = nsolute/1 kg of solvent nglucose = mglucose x 1 kg H2O = (0.67 mol/kg) x (0.1500 kg) = 0.10 mol MMglucose = massglucose /nglucose = 18.00 g /0.10 mol = 180 g/mol = [0.0 – (-23.3)] oC = 23.3 oC msolute = ΔTf/ Kf = (23.3 oC)/ (1.86 oC. kg/mol) = 12.5 mol/kg = nsolute/1 kg Solvent nglycol = mglycol x kgH2O = (12.5 mol/kg) x (10.0 kg) = 1.2 x 102 mol nglycol = massglycol /Mglycol massglycol = nglycol x Mglycol = (12.5 x 102 mol) x (62.1 g/mol) = 7.76 x 103 g = 7.76 kg = 0.24 oC mhormone = ΔTf/Kf = (0.24 oC)/ (5.12 oC. kg/mol) = 4.69 x 10-2 mol/kg = nhormone/1 kg Benzene nhormone = mhormone x 1 kg Benzene = (4.69 x 10-2 mol/kg) x (0.01500 kg) = 7.04 x 10-4 mol Mhormone = masshormone /nhormone = 0.546 g /7.04 x 10-4 mol = 776 g/mol Works cited Gregorian, Mike. "Chemistry." Socratic (2018): 15. Kottler, Jeffery. "Introduction to Chemistry." Lumen (2014): 44.

Name
Professor
Class
Date
Chemistry
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

1. 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = 46= 0.0217
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =

0.0217𝑚𝑜𝑙𝑒𝑠
0.101𝑙𝑖𝑡𝑟𝑒𝑠

= 0.215M
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑀𝑎𝑠𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 100%
1

= 101 ∗ 100 = 0.99%
𝑀𝑜𝑙𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
0.0217𝑚𝑜𝑙𝑒𝑠
=
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.215𝑚𝑜𝑙𝑒𝑠

= 0.1
𝑀𝑜𝑙𝑎𝑙𝑖𝑡𝑦 =
=

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒...

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Anonymous
Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

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