Answer: - 1.86°C
Explanation:
The depression of freezing points of solutions is a colligative property.
That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.
To solve the problem follow these steps:
Data:
Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol
2) Formulas
Tf = Normal Tf - ΔTf
ΔTf = i * kf * m
m = number of moles of solute / kg of solvent
number of moles of solute = mass in grams / molar mass
3) Solution
number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol
m = 0.2 mol / 0.5 kg = 1.0 m
ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C
Tf = 0°C - 1.86°C = - 1.86°C
Answer: - 1.86 °C
If you measure my hairline you will get about 90°
Homogeneous because it completely dissolved so its now one. Heterogeneous would be when they don't dissolve and you see 2 separate substances like sand in water.
Answer:
Molar solubility is 1.12x10⁻⁴M
Explanation:
The dissolution of magnesium hydroxide is:
Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻
The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:
Mg(OH)₂(s) ⇄ X + 2X
<em>Where X is solubility.</em>
<em />
If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:
2X = 2.24x10⁻⁴M
X = 2.24x10⁻⁴M/2
X =1.12x10⁻⁴M
<h3>Molar solubility is 1.12x10⁻⁴M</h3>
3
Let me know if you need/want an explanation
