8. Calculat e the total vapor pressure of a solution made by dissolving 25.0 g o

Question

8. Calculat e the total vapor pressure of a solution made by dissolving 25.0 g of glucose (CaH0%) in 215 g of water at 50 °C. MMcan2ox-1802 g/mol, MMipo-18.02 g/mol. Vapor Pressure of Ho@ 50°C-92,5 tom) (Answer: 91.4 torr) (similar Q.71 page 558) 398K 9. A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass of the unknown compound? A) 223 g/mo B294 g/mol C) 341 g/mol D)448 g/mol E) 195 g/mol Lysozyme is an enzyme used to cleave cell walls. lysozyme in 100.0 mL results in an osmotic pressure of 1.32 x 10 molar mass of lysozyme. A solution made by dissolving 0.0750 g of atm at 25 °C. Calculate the 10. (Answer: 1.39 x 10* g/mol.) (similar Q. 83)

Explanation / Answer

1. 1. Determine moles glucose and water.

moles glucose = 25.0 grams glucose x (1 mole glucose/180.2 grams glucose) = 0.139 moles glucose

moles H2O = 215.0 grams H2O x (1 mole H2O /18.02 grams H2O) = 11.93 moles H2O

2. Determine mole fraction of H2O.

mole fraction, x = moles H2O /total moles = 11.93 moles/(0.139 + 11.93)moles = 0.9885

3. Determine pressure of solution.

P = xP*

where:

P = pressure of solution

x = mole fraction of H2O = 0.9885

P* = pressure of H2O = 92.5 torr

P =(0.9885)*(92.5 torr) = 91.43 torr

Total pressure of solution is 91.43 torr.

2. Hint : osmotic pressure = (molarity)(ideal gas constant)(absolute temperature)

And molarity = mol/vol

And mol = given wt/molecular wt

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