1) The boiling point of water H 2 O is 100.0 °C at 1 atmosphere. A nonvolatile,

Question

1) The boiling point of water H2O is 100.0°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is glucose . How many grams of glucose, C6H12O6 (180.2 g/mol), must be dissolved in 249.0 grams of water to raise the boiling point by 0.450 °C ? ... g glucose.

2) The freezing point of ethanol CH3CH2OH is -117.30°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in ethanol is testosterone .
If 11.04 grams of testosterone, C19H28O2 (288.4 g/mol), are dissolved in 196.4 grams of ethanol ...
The molality of the solution is  m.
The freezing point of the solution is  °C.

3) The boiling point of ethanol, CH3CH2OH, is 78.500 °C at 1 atmosphere. Kb(ethanol) = 1.22 °C/m
In a laboratory experiment, students synthesized a new compound and found that when 11.04 grams of the compound were dissolved in 230.9 grams of ethanol, the solution began to boil at 78.818 °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

Explanation / Answer

1) elevation in boiling point

dTb = iKbm

with, Kb = 0.512 oC/m

we get,

0.450 = 1 x m x 0.512

molality of glucose needed (m) = 0.88 m

grams of glucose needed for solution = 0.88 x 0.249 x 180.156

                                                            = 39.476 g

2) molality of solution = 11.04 g/288.4 g/mol x 0.1964 kg = 0.195 m

Kf for ethanol = 1.99 oC/m

so,

change in freezing point of solution = dTf = iKfm = 1 x 1.99 x 0.195

                                                         = 0.388 oC

freezing point of solution = -117.30 - 0.388 = -117.69 oC

3) dTb = iKbm

(78.818 - 88.5) = 1 x 1.22 x 11.04/molar mass x 0.2309

molar mass of compound = 183.433 g/mol

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