± Molar Mass from Colligative Properties Part A The molar mass of a compound exp

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± Molar Mass from Colligative Properties Part A The molar mass of a compound expresses the ratio of mass to moles tert-Butyl alcohol is a solvent with a Kf of 9.10 C/m and a freezing point of 25.5 °C. When 0.807 g of an unknown colorless liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3° C molar massmass in grams number of moles Which of the following is most likely the identity of this unknown liquid? This quantity can be determined experimentally by accurately measuring the mass of the sample and determining the corresponding number of moles based on some property of the sample. Freezing point depression and osmotic pressure measurements are frequently determination. In each instance, the number of moles is determined from the colligative property of the solution. For osmotic pressure measurements, the number of moles is calculated from the volume of the solution and the molarity. In freezing point depression measurements, the number of moles is derived from the molality of the solution Hints O ethylene glycol (molar mass 62.07 g/mol) O 1-octanol (molar mass 130.22 g/mol) O glycerol (molar mass = 92.09 g/mol) O 2-pentanone (molar mass 86.13 g/mol) O 1-butanol (molar mass 74.12 g/mol) used for this type of Submit My Answers Give Up Part B A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.145 g of this subunit was dissolved in enough water to produce 2.00 mL of solution. At 28 C the osmotic pressure produced by the solution was 0.138 atm. What is the molar mass of the protein? Express your answer with the appropriate units Hints ValueUnits Submit My Answers Give Up Continue

Explanation / Answer

Part A.

molality = moles of solute/kg of solvent

moles = g/molar mass

solvent = 11.6 g = 0.0116 kg

solute = 0.807 g

Using formula,

dTf = mKf

where, m = molality of solution

(25.5 - 15.3) = m x 9.10

m = moles/0.0116 = 1.121 m

moles of solute = 0.013 mols

molar mass of solute = 62.07 g/mol

So, the correct answer would be, the unknown liquid is ethylene glycol (molar mass = 62.07 g/mol)

Part B.

osmotic pressure = MRT

where,

M = molarity of solution = moles/L of solution = g/molar mass x L of solution

volume of solution = 2 ml = 0.002 L

mass of solute = 0.145g

T = 28 oC = 28 + 273 = 301 K

Feed values,

M = 0.138/0.08206 x 301 = 5.587 x 10^-3 M

molar mass of the protein = 0.1145/5.587 x 10^-3 x 0.002

=10247.00 g/mol

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