freezing point of CoBr3 o Colorado Mesa University . OT x Search-l nesa... Muscl

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freezing point of CoBr3

o Colorado Mesa University . OT x Search-l nesa... Muscle Identification Share More> Home 2015 11:00 AM O 11.7/249/23/2015 11:13 PM Print | Calculator-d Periodic Table 2 of 8 era University Science Books prosented by Sapling Learning McQuarrie Peter A. Rock Ethan Gallogly Given that the vapor pressure of water is 17.54 Torr at 20 "C, calculate the vapor-pressure lowering of aqueous solutions that are 1.90 m in (a) sucrose. C12H22011, and (b) sodium chloride. Assume 100% dissociation for electrolytes. Number 11 (a) sucrose. C12H22011 /, = Torr Number (b) sodium chloride 1-11 Torr

Explanation / Answer

(1) According to Roults's law relative lowering of vapour pressure = mole fraction of solute.

                              (P0 - P) / Po = X

Where Po = vapour pressure of pure water = 17.54 torr

           P = vapour pressure of the solution

          X = mole fraction of solution = ( number of moles of solute / total number of moles )

Given that the molality of the solution , m = 1.90 m

Assuming the mass of water ( solvent) is 1.0 kg(=1000 g )

Molarmass of H2O is = (2x1) + 16 = 18 g/mol

Molar mass of sucrose,C12H22O11 = ( 12 x 12 ) + (22 x 1 ) + (11 x 16 ) = 342 g/mol

Molality , m = ( mass of solute / molar mass of solute ) / mass of solvent in kg

             1.90 = ( w / 342 ) / 1.0 )

                  w = 649.8 g

So number of moles of sucrose , n = mass / molar mass

                                                        = 649.8 / 342

                                                       = 1.9 moles

Number of moles of water (H2O) , n' = mass / molar mass

                                                             = 1000 g / 18 g/mol

                                                             = 55.56 mol

So mole fraction of sucrose , X = ( number of moles of solute / total number of moles )

                                                  = 1.90 / ( 1.90 + 55.56 )

                                                  = 0.033

Plug the values in Roult's law we get     (P0 - P) / Po = X

                                                               (P0 - P) / 17.54 = 0.033

                                                                (P0 - P) = 17.54 x 0.033

                                                                              = 0.58

Therefore the lowering of vapour pressure is = (P0 - P) = 0.58 Torr

(2) According to Roults's law relative lowering of vapour pressure = mole fraction of solute.

                              (P0 - P) / Po = X

Where Po = vapour pressure of pure water = 17.54 torr

           P = vapour pressure of the solution

          X = mole fraction of solution = ( number of moles of solute / total number of moles )

Given that the molality of the solution , m = 1.90 m

Assuming the mass of water ( solvent) is 1.0 kg(=1000 g )

Molarmass of H2O is = (2x1) + 16 = 18 g/mol

Molar mass of sodium chloride, NaCl is = 23+35.5 = 58.5 g/mol

Molality , m = ( mass of solute / molar mass of solute ) / mass of solvent in kg

             1.90 = ( w / 58.5 ) / 1.0 )

                  w = 111.15 g

So number of moles of sodium chloride , n = mass / molar mass

                                                                    = 111.15 g / 58.5 g/mol

                                                                    = 1.9 mole

Number of moles of water (H2O) , n' = mass / molar mass

                                                             = 1000 g / 18 g/mol

                                                             = 55.56 mol

So mole fraction of sucrose , X = ( number of moles of solute / total number of moles )

                                                  = 1.90 / ( 1.90 + 55.56 )

                                                  = 0.033

Plug the values in Roult's law we get     (P0 - P) / Po = i x X   

             where i = 2 since upon dissolution NaCl produces 2 moles of ions vai Na+ & Cl-

                                                               (P0 - P) / 17.54 = 2 x 0.033

                                                                (P0 - P) = 17.54 x 2x 0.033

                                                                              = 1.16

Therefore the lowering of vapour pressure of sodium chloride solution is = (P0 - P) = 1.16 Torr

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