Asked by Anonymous
Camphor melts at 179.8C and has a Kf = 40.0 C/m. When 0.186 g of an unknown nonvolatile nonelectrolyte is dissolved in 22.01 g of liquid camphor, the freezing point is found to be 176.7C. What is the molar mass of this solute?
Answers
Answered by
GothTutor
179.8 − 176.7 = 3.1 °C
Δt = i Kf m
3.1 °C = (1) (40.0 °C kg mol¯1) (x / 0.02201 kg)
3.1 °C = (1) (1817.356 °C mol¯1) (x)
x = 0.001705775 mol
0.186 g / 0.001705775 mol = 109 g/mol
Δt = i Kf m
3.1 °C = (1) (40.0 °C kg mol¯1) (x / 0.02201 kg)
3.1 °C = (1) (1817.356 °C mol¯1) (x)
x = 0.001705775 mol
0.186 g / 0.001705775 mol = 109 g/mol
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