The freezing point of water is 0.00°C at 1 atmosphere. If 10.66 grams of aluminum bromide, (266.7 g/mol), are dissolved in 225.9 grams of?
Update:
The freezing point of water is 0.00°C at 1 atmosphere.
If 10.66 grams of aluminum bromide, (266.7 g/mol), are dissolved in 225.9 grams of water
The molality of the solution is
m.
The freezing point of the solution is
°C.
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Answers & Comments
(10.66 g AlBr3) / (266.7 g AlBr3/mol) / (0.2259 kg) = 0.17694 mol/kg = 0.177 m
(0.17694 m) x (1.86 °C/m) = 0.329°C change
0.00°C - 0.329°C = -0.33°C