an..,. X t values if needed for this question. The boiling point of chloroform,

Question

an..,. X t values if needed for this question. The boiling point of chloroform, CHCls, is 61.700 °C at 1 atmosphere. K (chloroform)-3,67 °C/m In a laboratory experiment, students synthesized a new compound and found that when 12.20 grams of the compound were dissolved in 262.5 grams of chloroform, the solution began to boil at 62.291 °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 9 more group attempts remaining

Explanation / Answer

Solution-

DT = Kb * m

Here DT is the change in the boiling point of the pure solvent compared to the boiling point of the solution,

Kb is the ebulloscopic constant and m the molality of the solution.

Plugging this equation with the given data we arrive to

62.691°C – 61.700°C =0.991 °C/m * m

(0.991 °C / 3.67°C)* m = m

0.2700 m

Now

Moles of unknown /Kg of Chloroform = molality of solution

As moles of unknown can be calculated from

Moles of unknown = mass of unknown / MM unknown

Now by Combining both equation we get

(Mass of unknown / MM)/Kg of chloroform = 0.2700 m
(12.20/ MM)/0.2700 Kg) = 0.2700

12.20 / MM = 0.2700*0.2224

12.20/MM = 0.060048
12.20/0.060048 = MM

MM =203.17 g/mol
Answer

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