What is the expected change in the freezing point of water for a 0.015 m aqueous solution of alcl3?

- 0.028 °C

Explanation:

The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: ΔTf = Kf.m.

Where, ΔTf is the depression in water freezing point (ΔTf = ???°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of AlCl₃ (m = 0.015 m).

∴ ΔTf = Kf.m = (1.86 °C/m)(0.015 m) = 0.0279 °C ≅ 0.028 °C.

∵ ΔTf = freezing point of pure water - freezing point of the solution in presence of AlCl₃.

∴  freezing point of the solution in presence of AlCl₃ = freezing point of pure water - ΔTf = 0.0 °C - 0.028 °C = - 0.028 °C.

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom

Sorry i think the answer is b