Please show all work 25. Pure ethylene glycol, HOCH2CH2OH, is added to 2.00 kg o
ID: 703848 • Letter: P
Question
Please show all work
25. Pure ethylene glycol, HOCH2CH2OH, is added to 2.00 kg of water in the cooling system of a car. The vapor pressure of the water in the system when the temperature is 90 ° C is 457 mm Hg. What mass of glycol was added? (Assume the solution is ideal. See Appendix G for the vapor pressure of water.) 30, A solution of glycerol, C3H5(OH);' in 735 g of water has a boiling point of 104.4 C at a pressure of 760 mm Hg. What is the mass of glycerol in the solution? What is the mole fraction of the solute? Freezing Point Depression 31, A mixture of ethanol, C2H5OH. ?nd water has a freezing point of-16.0?C. (?) What is the molality of the alcohol? (b) What is the weight percent of alcohol in the solution? 41. If52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? ( Assume the van't Hoff factor, i, for LiF is 2.) 52. (a) Which aqueous solution is expected to have the higher boiling point: 0.10 m Na2SO or 0.15 m sugar? (b) For which aqueous solution is the vapor pressure of water higher: 0.30 m NH4NO3 or 0.15 m Na2SO4'? 61. (a) Which solution is expected to have the higher boiling point: 0.20 m KBr or 0.30 m sugar? (b) Which aqueous solution has the lower freezing point: 0.12 m NH4NO or 0.10 m Na-CO,?Explanation / Answer
Ans 25
Vapor pressure of pure water at 90 C = 525.8 mmHg
From the Raoult’s law
Vapor pressure of water in system = mole fraction of water x vapour pressure of pure water
457 mmHg = mole fraction of water x 525.8 mmHg
mole fraction of water = 0.8692
Moles of water = mass/molecular weight
= 2000g / 18g/mol
= 111.11 mol
Total Moles in the system = moles of water / mol fraction of water
= 111.11 / 0.8692
= 127.83 mol
moles of ethylene glycol = total moles - moles of water
= 127.83 - 111.11
= 16.72 mol
Mass of ethylene glycol added = moles x molecular weight
= 16.72 mol x 62.068
= 1037.86 g
= 1.038 kg
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