Explanation:
Let the volume of the solution be 100 ml.
As the volume of glycol = 50 = volume of water
Hence, the number of moles of glycol = 
= 
= 
= 0.894 mol
Hence, number of moles of water = 
= 2.77
As glycol is dissolved in water.
So, the molality = 
= 17.9
Therefore, the expected freezing point = 
= 
Thus, we can conclude that the expected freezing point is
.
I would say G sorry if it’s not right
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:
We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.
Explanation:
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No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .