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A glucose solution contains 50.5 of glucose in 475 of water. Compute the freezing point and boiling point of the solution Assume a density of 1.00 for water?
Wiki User
∙ 13y ago
50.5g of glucose
glucose is 180.18 g/mol
50.5/180.18 = 0.280 mol glucose
0.280 mol glucose/0.475 kg H2O = 0.589 m which is molality
deltaTsubF = (-1.86 degress C/m)(0.589) = -1.10 degrees C
-1.10 degrees C + 0.00 degrees C = -1.10 degrees C as your freezing point seeing that 0.00 degrees is the standard freezing point of water
deltaTsubB = (0.512 degrees C/m)(0.589) = 0.301568
it asks for 6 sig figs so
0.301568 degrees C + 100.000 degrees C (boiling point H2O) = 100.302 degrees C
Mastering Chemistry sucks sometimes....
Wiki User
∙ 13y ago
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