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

Question

Use the References to access important values if needed for this question The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(wter) 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.84 grams of the compound were dissolved in 241.0 grams of water, the solution began to boil at 100.525 °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 Submit Answer Retry Entire Group 8 more group attempts remaining

Explanation / Answer

DTb = Tb-T0

i = vanthoff factor of solute = 1

T0 = boiling point of solvent(water) = 100 C

Tb = boiling point of solution = 100.525 C

kb for water = 0.512°C·kg/mol

m = molality of solution = (w/M)*1000/W

w = Amount of compound = 14.84 g

M = molarmass of compound = x g/mol

W = Amount of water = 241.0 g

DTb = i*Kb*(w/M)*(1000/W)

100.525-100 = 1*0.512*(14.84/x)*(1000/241.0)

x = 60.05 g/mol

molarmass of compound = 60.05 g/mol

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