The freezing point of benzene , C 6 H 6 , is 5.500 ° C at 1 atmosphere. K f ( be

Question

The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = -5.12 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 12.47 grams of the compound were dissolved in 296.9 grams of benzene, the solution began to freeze at 4.754 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound ?

? g/mol

The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 11.21 grams of the compound were dissolved in 259.6 grams of diethyl ether, the solution began to boil at 34.598 °C. The compound was also found to be nonvolatile and a non-electrolyte.


What is the molecular weight they determined for this compound ?

? g/mol

Explanation / Answer

Answer –

1) We are given, Freezing point of benzene = 5.500oC , Kf for benzene = 5.12 °C/m
mass of compound = 12.47 g

Mass of benzene = 296.9 g , Freezing point of solution = 4.754oC

We are given mass of unknown substance and we need to calculate molar mass of unknown substance, so we need to calculate first moles of unknown substance and for calculating moles we need to calculate molality

We know formula for freezing point depression,

Tf = Kf* m

Tf = 5.500oC – 4.754oC

       = 0.746oC

So, m = Tf / Kf

          = 0.756 oC / 5.12 oC.m-1

          = 0.146 m

We know molality means mole of solute per kg of solvent

So, molality = mole/kg of solvent

Mass of solvent = 296.9 g = 0.2969 kg

Moles of solute = molality * kg of solvent

                         = 0.146 m * 0.2969 kg

                         = 0.0433 moles of solute

Now we calculated moles of unknown substance and we are given mass of unknown substance,

So, molar mass of unknown substance = mass / moles

                                                              =12.47 g/ 0.0433 mole

                                                         = 288.26 g/mole

So, molar mass of unknown substance is 288.26 g/mol

2) We are given, Boling point of diethyl ether = 34.500oC , Kb for diethyl ether = 2.02 °C/m
mass of compound = 11.21 g

Mass of benzene = 259.6 g , Boling point of solution = 34.598oC

We are given mass of unknown substance and we need to calculate molar mass of unknown substance, so we need to calculate first moles of unknown substance and for calculating moles we need to calculate molality

We know formula for boiling point elevation,

Tb = Kb* m

Tb = 35.500oC – 34.598oC

       = 0.902oC

So, m = Tb / Kb

          = 0.902 oC / 202 oC.m-1

          = 0.446 m

We know molality means mole of solute per kg of solvent

So, molality = mole/kg of solvent

Mass of solvent = 259.6 g = 0.2596 kg

Moles of solute = molality * kg of solvent

                       = 0.446 m * 0.2596 kg

                         = 0.116 moles of solute

Now we calculated moles of unknown substance and we are given mass of unknown substance,

So, molar mass of unknown substance = mass / moles

                                                              =11.21 g/ 0.116 mole

                                                         = 96.70 g/mole

So, molar mass of unknown substance is 96.70 g/mol

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