Use the References to access important values if needed for this question. The f

Question

Use the References to access important values if needed for this question. 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 10.80 grams of the compound were dissolved in 225.9 grams of ethanol, the solution began to freeze at -117.406 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

Explanation / Answer

depression in freezing point = -117.3 -(-117.406)

depression in freezing point = 0.106 C

now

dTf = Kf x m

0.106 = 1.99 x m

m = 0.053266

now

moles of solute = mass of ethanol (kg) x molality

= 0.2259 kg x 0.053266 mol/kg

= 0.012033 mol

now

molar mass of solute = mass of solute / moles of solute

= 10.8 g / 0.012033 mol

= 897.54 g/mol

so

molecular weight of the compound is 897.54 g/mol

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.