A 0.102 g sample of an unknown compound was dissolved in 20.00 g of cyclohexane.

Question

A 0.102 g sample of an unknown compound was dissolved in 20.00 g of cyclohexane. The resulting solution was found to have a freezing point of 5.63 degreeC. The freezing point of pure cyclohexane is 6.55 degreeC and its freezing point depression constant (Kf) is 20.0 degreeC/m. What is the molecular weight of the unknown compound? (Assume the compound is non-volatile and does not ionize in the solution.) 85.9 g/mol 111 g/mol 137 g/mol 186 g/mol 245 g/mol A 1.97 g sample of a protein having a molecularweight of 82100 g/mol was dissolved in enough water to produce 41.50 mL of solution. What is the osmotic pressure of this solution at 25 degreeC? Note that the gas constant R has a value of 0.082057 L atm /K mol. You may also find the following equations useful:K = degreeC + 273.15 1 atm = 760torr 1.63 torr 3.31 torr 4.92 torr 7.95 torr 10.8 torr A 2.26 g sample of a protein of unknown molecular weight was dissolved in enough water to produce 21.3 mL of solution. The solution was

Explanation / Answer

1)
DTf = Kf*molality
(6.55-5.63) = Kf*molality

   0.92 = 20*(0.102/mol.wt)*1000/20
   mol.wt = 111 gm/mol
2)
   Pi= conc.*R*T
   Pi = 1.97/82100)*(1000/41.5)*0.082057*298
Pi = 0.0141 atm
as given
1atm. ...............760 torr
0.0141 atm ..........?
Pi(osmotic pressure) = 10.8 torr

3)
760 torr............1atm
2.45 torr............?
= 0.00322 atm
so we Know that
Pi = c*R*T
0.00322 = (2.26/mol.wt)*(1000/21.3)*0.082057*298
mol. wt = 8.05*10^5 gm /mol

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