15.4-g sample of a nonelectrolyte is dissolved in 174.0 g of water the solution will freeze at –0.93°C.the molar mass of the nonelectrolyte?
Update:
kf for water is 1.858 degree celsius/m
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kf for water is 1.858 degree celsius/m
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Calculate the molality of the solution:
Change in freezing point = -Kf m
-0.93 C = 1.858 C/m (m)
m = 0.500 molal solution
0.500 molal = moles solute / 0.174 kg water
moles solute = 0.0876
molar mass solute = 15.4 g / 0.0876 mol = 176 g/mol