Question

tert-Butyl alcohol is a solvent with a Kf of 9.10 °C/m and a freezing point of 25.5 °C.w liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3 °C.Which of t Tiquid? View Available Hint(s) ethylene glycol (molar mass = 62.07 g/mol) 1-octanol (molar mass = 130.22 g/mol) glycerol (molar mass = 92.09 g/mol) 2-pentanone (molar mass = 86.13 g/mol) 1-butanol (molar mass = 74.12 g/mol)

Answer 1

Formula to calculate depression in freezing point is as follows: AT; К.-b-i = Freezing point of a solvent tert-butyl alcohol is 25.5°C , and freezing point for the solution is 15.3°C .Calculate the change in freezing point is as follows: AT, T (solvent)-T, (solution) = 25.5°C-15.3°C = 10.2°C Calculate the molarity of the solution from depression in freezing point is as follows: AT; (For organic solvents i is 1) К, b= 10.2 b = 9.10 = 1.12 mol/kg Calculate the number of moles of solute is as follows: 1 kg mol x(11.6 g)x kg Number of moles of soute = |1.12 1000 g = 0.0013 mol

Calculate the molecular weight of solute from number of moles is as follows: Weight of solute Molecular weight of solute Number of moles of solute 0.807 g 0.013 moles Molecular weight of solute 0.807 g Molecular weight of solute 0.013 moles 62.07 g/mole 11 Therefore, liquid with molecular mass 62.07 g/mole is ethylene glycol. Hence the solute is ethylene glycol