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
Answer:
315.51g/mol
Explanation:
137(33 + (16.00 + 1.01) 2 + 8 [1.01 (2) + 16.00] = 315.51g/mol
Answer:
put a salt into the beakers
Answer:
Following are the solution to the given points:
Explanation:
In point a, the answer is only ions because of 
has a strong electrolyte.
In point b, the answer is the only molecules because of ethanol 
, it has a nonelectrolyte.
In point c, the answer is the few ions because of hydrocyanic acid HCN, which has a weak electrolyte.
