Need help with molality and van \'t hoff 22.6500g FeCl3* 6 H20_are dissolved in

Question

Need help with molality and van 't hoff 22.6500g FeCl3* 6 H20_are dissolved in 15.3575 g water. The final volume of the solution is 29.75 mL. The solution is determined to freeze at -8 985 degree C P water freezes at +0.100 degree C on the same thermometer. Calculate the molality of the solution. (4) Determine the experimental value of the van't Hoff factor. (3) Calculate the apparent per cent dissociation of the salt in this solution.(5) A 0.2765 m solution of the same salt is demonstrated to be 62.3% dissociated. At what temperature, on this thermometer, is this solution expected to freeze? (5)

Explanation / Answer

a)

molality = mol/kg solvent

mol = MW = 162.2 +6*18 = 270.2

mol = 22.65/270.2 = 0.0838

kg solvent = 15.3575 = 15.357/1000

molal = 0.0838/(15.357/1000 = 5.4567 mol per kg

b)

dTf = Kf*m*i

-8.985 = (1.86)(5.4567)(i

i = 8.985 /((1.86)(5.4567)) = 0.8852

c)

apparent %

0.8852 disosciation

d)

if m = 0.2765 molal

62.2% dissocaition

T is bellow, then expect this NOT to freeze

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