Use the van\'t Hoff factors in the table below to calculate each colligative pro

Question

Use the van't Hoff factors in the table below to calculate each colligative property Table. Van't Hoff Factors at 0.05 m Concentration in Aqueous Solution i Expected i Measured Solute 1 1 Nonelectrolyte 2 19 NaCl 2 MgSO4 1.3 3 MgCl2 2.7 3 K2SO4 2.6 4 FeCl3 3.4 Part A the melting point of a 0.150 m iron chloride solution (ll) Express your answer using three significant figures. Th 100.26 Submit My Answers Give Up incorrect, Try Again; 3 attempts remaining Part B the osmotic pressure of a 0.084 M potassium sulfate solution at 298 K Express your answer using two significant figures atm Submit My Answers Give Up

Explanation / Answer

A)

Apply:

dTb = Kb*m*i

where, dTF = increase in BP; Kb = water's constant ad i = number of ions in solution FeCl3 = Fe+3 + 3Cl-

so

dTb = 0.512*0.15*(3.4) = 0.2611

So

Tboiling = 100 + 0.2611= 100.2611°C --> 3 sig fig = 100.°C

B)

The formula:

PI = M*RT*i

For K2SO4 = 2.6

M = 0.084 M and T = 298 K and = 0.082 L/molK

so

PI = 0.084*0.082*298*2.6 = 5.336 atm = 5.34 atm (3sig fig)

C)

BP of 1.12% mass MgCl2

so..

dTb = Kb*m*i

i = Mg+2 and 2cl- = 1+2 = 3 ions (1.3 experimental)

so

dTb = 0.512*m *1.3

m = mol of solute / kg solvetn

assume a basis of 100 g of mix:

1.12/100*100 = 1.12 g of MgCl2

mol = mass/MW = 1.12/95.211 = 0.011763 mol of MgCl2

kg solvent = (1-0.0112)*100 = 98.88g = 0.09888 kg

so

molality =0.011763/0.09888 = 0.1189

so

dTb = 0.512*0.1189 *1.3 = 0.079139°C

Tb = 100 + 0.079139= 100.079139 °C = 3 sig fig 100.

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