1a. The freezing point of ethanol , CH 3 CH 2 OH , is -117.300 ° C at 1 atmosphe

Question

1a. The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere.  Kf(ethanol) = 1.99 °C/m
In a laboratory experiment, students synthesized a new compound and found that when 10.62 grams of the compound were dissolved in 208.3 grams of ethanol, the solution began to freeze at -117.854 °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)

1b. The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 11.59 grams of the compound were dissolved in 246.9 grams of water, the solution began to boil at 100.387 °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

Hence m= 0.554/1.99= 0.2784

Molality, m= moles of solute/ kg of solvent = moles of solute/0.2083

Moles of solute =0.2083*0.2784= 0.058

Moles= mass/molar mass, molar mass= mass/moles= 10.62/0.588=183 g/mole

2.

Boiling point elevation = i*kb*m, i=1 for non volatile solute,

m= boiling point elevation/ kb*i= (100.387-100)/0.5*1=0.774

Molality= moles of solute/ kg of solvent = moles of solute/0.2469

Moles of solute= 0.2469*0.774= 0.19

Moles= mass/molar mass, molar mass= mass/moles= 11.59/0.19 =61 g/mole

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