Calculate the freezing of solution of 3.46 g of compound,X, in 160 g of benzen.when separated sample of X was vaporised, it density 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzene is 5.45 c and Kf is 5.12 c/m.

Answer 1

Use the density and PV = nRT to solve for molar mass of X.
Then 3.46g X/molar mass X = moles.

molality = moles/kg solvent

delta T = Kf*molality

normal freezing point - delta T = new freezing point.

Answer 2

1.48

Answer 3

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Answer 4

calculat the freezing point of a solution of 3.46g of a compound,X , in 160g of benzene. when a separate sample of X was vaporised, its density was found to be 3.27g/L at 116C and 773 torr. the freezing point of pure benzene is 5.45C, and Kf is 5.12Ckg/mol

Answer 5

a solution of crab hemocyani, a pigimented protein extracted from crabs, was prepared by dissolving 0.750g of the protein in 125mL of mater. at 4C, a 2.6mmHg osmotic-pressure rise of the solution was observed. the solution had a density of 1.00g/mL.determine the molar mass of the protei.

Answer 6

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Answer 10

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Answer 11

Calculate the freezing point of a solution of 3.46g of comoundx in 160g of benzene, when it density found to be 3.27g/L at 116c and 773 torr. The freezing point of pure benzene is 5.45(kf is 5.12kg.mol)determine .a,the concentration solution.b,the number of mole in the solution.c, molality. c,freezing point depression