This is a question about the colligative property known as freezing point depression. Freezing point depression (the amount the normal freezing point of the solvent is decreased) can be calculated with this equation:
ΔT = i Kf<span> m
</span>
Where i (the van't Hoff factor) is the degree of dissociation of the solute, Kf is the freezing point depression constant, and m is the molality of the solution.
Here i = 2 (KCl dissociates into 2 ions, K+ and Cl-), Kf = 1.86 C/m (for water), and m = 0.743m).
ΔT = 2 x 1.86 C/m x 0.743m = <span>2.764C
</span>
That means the freezing point of the solution is 2.764C less than the pure solvent (water), making it 0C - 2.764C = -2.764C.
Answer: the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g
Explanation:
1. <u>Preliminars</u>;
- Molar mass of CH₃ OH: 32.04 g/mol
- Molar mass of C₂H₅OH: 46.07 g/mol
- Molar mass of CO₂: 44.01 g/mol
- Molar mass equation: molar mass = mass in grams / number of moles
2. <u><em>Algebraic equation # 1: mass of sample burned in terms of each reagent.</em></u>
3. <u><em>Algebraic equation # 2: mass of carbon dioxide produced in terms of each reagent</em></u>.
a) Combustion of methyl alcohol
- 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
- 2 × 32.04 g CH₃OH → 2 × 44.01 g CO₂
- Mass of CO₂ produced from M grams of CH₃OH = M / 32.04 × 44.01 = 1.374 M
b) Combustion of ethyl alcohol
- C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)
- 46.07 g CH₃OH → 2 × 44.01 g CO₂
- Mass of CO₂ produced from E grams of C₂H₅OH = E / 46.07 × 88.02 = 1.911 E
c) Mass of CO₂ (g) obtained, in terms of each reagent
- 1.374 M + 1.911 E = 0.386 ----- (2)
4) <u>System of equations</u>
- 1.374 M + 1.911 E = 0.386 ----- (2)
5) <u>Solution</u>
- From equation (1): E = 0.220 - M
- Substituting in equation (2): 1.374 M + 1.911 (0.220 - M) = 0.386
- 1.374 M + 0.4204 - 1.911 M = 0.386
- M = 0.0344 / 0.537 = 0.0641 g
Answer: the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g
In the isotope of Si-28, there are 14 protons.
C, Electronegativity Increases