133.3) is added to 100.0 g of water. Find the molality of la.) 13.33 g of AlCl,

Question

133.3) is added to 100.0 g of water. Find the molality of la.) 13.33 g of AlCl, (Mw (8) the solution. (8) b.) Find the freezing point of this solution. Kr of H2O 1.86 oc. Hint: Remember that AlCl3 forms an ionic solution, and assume 100% ionization (8) c) Calculate the mole fraction of water. (5) d.) The vapor pressure of pure water at 25 0C 23.78 torr. Find the vapor pressure of this AlCl, solution at the same temperature. (8) e.) Assume that the density of this AlCl3 solution 1.04 g/ml. Find the molarity of this solution.

Explanation / Answer

molality= moles of solute/ kg of solvent

moles of solute= mass/molar mass= 13.33/133.3=0.1 moles

mass of water= 100gm= 100/1000 kg= 0.1

Molarity, m= 0.1 moles/0.1= 1M

2. depression in freezing point= i*kf*m, i= Van;t Hoff factor for AlCl3 ( AlCl3------>Al+3+3Cl-, i=1(Al+3)+3(Cl-)= 4

depression in freezing point= 4*1.86*1= 7.44

freezing point = freezing point of water- freezing point depression= 0-7.44= -7.44

3. moles of water= 100/18= 5.55, total moles= moles of water+ moles of AlCl3= 5.55+0.1=5.65

mole fraction of water= moles of water/ total moles= 5.55/5.65= 0.982

4.vapor pressure of water in solution = mole fraction of water* pure component vapor pressure = 0.982*23.78 Torr=23.35 Torr

5. volume of solution = mass/ densty= (100+13.33)/1.04 =108.9 ml= 108.9/1000 =0.1089 L

molarity= moles of solute/ liter of solution = 0.1/0.1089 M=0.9182 M

6.

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