Pure benzophenone freezes at 48.1 degree C. A solution of 1.25 g of biphenyl (C_

Question

Pure benzophenone freezes at 48.1 degree C. A solution of 1.25 g of biphenyl (C_12H_10; molar mass = 154.2 g/mol) plus 25.38 g of benzophenone froze at 16.8 degree C. Calculate the K_f of benzophenone. A mixture of 1.03 g of unknown plus 23.94 g of benzophenone had the following cooling behavior: Prepare a graph showing temperature on the ordinate and time on the abscissa. Label the point at which crystals first appeared, and identify the freezing temperature of the solution. Using the data for benzophenone from Problem 6, calculate the molality of the solution, then the molar mass of the unknown.

Explanation / Answer

Solution:- (6) Pure benzophenone(solvent) freezes at 48.1 degree C.

mass of biphenyl(solute) = 1.25g

moles of solute = 1.25g x 1mol/154.2g = 0.00811 mol

mass of solvent = 25.38g = 0.02538 kg

molality of the solution = mol of solute/kg of solvent = 0.00811 mol/0.02538kg = 0.3195 mol/kg

freezing point of solution = 16.8 degree C.

depression in freezing point, delta Tf = 48.1 - 16.8 = 31.3 0C

we know that,

delta Tf = m kf

where kf is molal freezing point constant and m is molality.

Let's plug in the values...

31.3 0C = (0.3195 mol/kg) kf

kf = 31.3 0C/0.3195 mol/kg = 97.97 0C. kg/mol

(7) (a) We take temperature along y-axis and time along x-axis and plot a graph. we add the points to get the line and then draw a line from x-axis (10.9 min) towards the curve that we get on join the points. Where this line cut the curve we draw another line from that point and see where it cuts y-axis. This is the freezing point of the solution and we get this value as 12.6 degree C.

(b) delta Tf = 48.1 - 12.6 = 35.5 0C

kf = 97.97 0C. kg/mol

delta Tf = m. kf

m = delta Tf/kf

let's plug in the values...

m = 35.5 0C/ 97.97 0C. kg/mol = 0.362 mol/kg

we know that molality is moles of solute/kg of solvent.

mass of solvent used is 23.94g that is 0.02394 kg. Now we could calculate the moles of solute.

0.02394 kg x 0.362 mol/kg = 0.00867 mol

mass of unknown solute used is 1.03g.

So, molar mass of unknown solute = 1.03g/0.00867 mol = 118.8 g/mol

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