1) The freezing point of water is 0.00 °C at 1 atmosphere. If 11.04 grams of alu

Question

1) The freezing point of water is 0.00°C at 1 atmosphere.
If 11.04 grams of aluminum chloride, (133.3 g/mol), are dissolved in 196.4 grams of water.
The molality of the solution is  m.
The freezing point of the solution is  °C.

2) The boiling point of water is 100.0°C at 1 atmosphere. (Kb(water) = 0.512°C/m)   
How many grams of barium bromide (297.1 g/mol), must be dissolved in 249.0 grams of water to raise the boiling point by 0.450°C ? g barium bromide.

3) The boiling point of water/H2O is 100.00 °C at 1 atmosphere.
If 14.23 grams of aluminum acetate, (204.1 g/mol), are dissolved in 293.7 grams of water ...
The molality of the solution is  m.
The boiling point of the solution is  °C.

Explanation / Answer

The molality is the moles of AlCl3 per kilogram of water

Molecular weight of aluminium chloride =133.3 g

Mass of AlCl3 present in solution= 11.04 g/mol

Therefore, no. of moles of AlCl3 present in solution= 11.04 /133.3 = 0.08 moles.

No. of moles of aluminium chloride present in 196.4g of water= 0.08 moles

Therefore, no. of moles of AlCl3 in 1000g of water: m = (0.08 ×1000)/ 196.4= 0.42 m

depression in freezing point DT = i x k(f) x m

where DT = change in temperature, i = van't Hoff factor, k(f) = freezing point depression constant, m = molality

AlCl3 dissociates as

AlCl3Al3+ + 3Cl-

Assuming complete dissociation, i=4

DT = 4 x kf x 0.42 = x (kf value is not given)

New freezing point will be 0-x

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