For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Explanation:
an increase in concentration increases the rate of the reaction. This is because there are more reactant particles available which allows for more effective collisions between reactant particles in a given period of time. More effective collisions bring about a faster rate of reaction.
One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.
pV = nRT
p = nRT/V
p= 1 x 0.08205 x 1000/ 2
p = 41.025 Pa
Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.
