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posted by  Jordan1331 on 9/9/2008 12:48:43 PM  |  status: Live  

Please help, al points given

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Given that you have 625 g of H2O in a 0.0245 m AgNO3 solution, calculate the mole fraction and weight % of AgNO3.

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posted by Jesley on 9/10/2008 1:24:51 AM  |  status: Live
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Jordan1331's comment:
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molality of the solution = 0.0245m
so, there is  0.0245moles of solute in 1kg of solvent, the number of moles of solute in 625g or
0.625kg  of solvent is
      moles of solute(AgNO3) = (0.625kg)*(0.0245mol/1kg)
                           =0.0153mol
mass of solvent, water = 625g
moles of solvent = 625g/18g/mol = 34.72mol
molar mass of AgNO3 = 107.9+14.01+3(16) =169.91g/mol
mass of solute AgNO3 = moles *molar mass
                                   = 0.0153mol*169.91g/mol =2.6g
mole fraction of solute = moles of solute/total moles(solute+solvent)
                                     = (0.0153mol)/(0.0153mol+34.72mol)
                                         =4.404x10-4
weight percent of AgNO3  = mass of AgNO3 /total mass *100
                                               =(2.6g)/(2.6+625g) *100
                                                 =0.414%
 
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