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0.60 grams of a non electrolyte was dissolved in a total of 12.0 mL of water. The osmotic pressure of the solution was measured to be 7.55 atm at 25 °C. Determine the molecular mass of this non electrolyte. Possible useful information: . . . 0 R = 0.0821 Latm/K mol 07= MRT • PV = nRT

0.60 grams of a non electrolyte was dissolved in a total of 12.0 mL of water. The osmotic pressure of the solution was measured to be 7.55 atm at 25 °C. Determine the molecular mass of this non electrolyte. Possible useful information: . . . 0 R = 0.0821 Latm/K mol 07= MRT • PV = nRT

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter12: Solutions
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Problem 12.96QP: A sample of aluminum sulfate 18-hydrate, Al2(SO4)3. 18H2O, containing 125.0 mg is dissolved in 1.000...
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0.60 grams of a non electrolyte was dissolved in a total of 12.0 mL of water. • The osmotic pressure of the solution was measured to be 7.55 atm at 25 °C. . Determine the molecular mass of this non electrolyte. • Possible useful information: R 0.0821 L atm/K mol R . • π = MRT • PV = nRT Psoln Posolvent Xsolvent O 162 g/mol 33.0 g/mol O 28.0 g/mol 81.0 g/mol 56.0 g/mol
Transcribed Image Text:0.60 grams of a non electrolyte was dissolved in a total of 12.0 mL of water. • The osmotic pressure of the solution was measured to be 7.55 atm at 25 °C. . Determine the molecular mass of this non electrolyte. • Possible useful information: R 0.0821 L atm/K mol R . • π = MRT • PV = nRT Psoln Posolvent Xsolvent O 162 g/mol 33.0 g/mol O 28.0 g/mol 81.0 g/mol 56.0 g/mol
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