PART A A solution is prepared from 4.5730 g of magnesium chloride and 43.232 g o

Question

PART A

A solution is prepared from 4.5730 g of magnesium chloride and 43.232 g of water. The vapor pressure of water above this solution is found to be 0.3624 atm at 348.0 K. The vapor pressure of pure water at this temperature is 0.3804 atm.

Find the value of the van't Hoff factor i for magnesium chloride in this solution.

PART B

Use the van't Hoff factors in the table below to calculate each colligative property.

-the melting point of a 6.00×102m iron(III) chloride solution

-the osmotic pressure of a 0.087 M potassium sulfate solution at 298 K

-the boiling point of a 1.25 % by mass magnesium chloride solution

Table. Van't Hoff Factors at 0.05 m Concentration in ueous Solution i Expected i Measured Solute Nonelectrolyte 1 1 2 NaCl 1.9 2 MgSO4 1.3 3 MgCl2 2.7 3 K2 SO4 2.6 4 FeCl3 3.4

Explanation / Answer

(A) Observed realtive lowering of vapour pressure = (Po - P) /Po = (0.3804 - 0.3624) / 0.3804 = 0.047

Po = pressure of pure solvent, P = partial pressure of solvent in water.

Calculated relative lowering of vapour pressure (Po - P) /Po ) = xMgCl2

xMgCl2 = mole fraction of MgCl = nMgCl2 / (nMgCl + nH2O)

nMgCl2 = moles of MgCl2 = mass / molar mass = 4.5730 / 95.0 = 0.048 moles

nH2O = 43.232 / 18.0 = 2.40 moles

xMgCl2 = nMgCl2 / (nMgCl + nH2O) = 0.048 / (0.048 + 2.4) = 0.0196

Hence calculated relative lowering of vapour pressure (Po - P) /Po = 0.0196

vant Hoff's factor i = experimental (observed) colligative property/  caculated colligative property = 0.047/0.0196 = 2.39

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