The addictive quality is a reduction in vapour pressure.
The vapour pressure at a liquid's normal boiling point is the same as the ordinary atmospheric pressure, which is 1 atmosphere, 760 Torr, 101.325 kPa, or 14.69595 psi.
The pressure that results from liquids evaporating is known as vapour pressure. Surface area, intermolecular forces, and temperature are three often occurring variables that affect vapour press.
lower vapour pressure
raising the boiling point
Low-temperature depression
Osmotic force
They are all dependent on the solute; when you increase the solute, the colligative property and the ratio you added may change.
The Van't Hoff Factor is another option to examine (i). the number of dissolved ions. The colligative property will be further altered if the solute is ionic.
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just K Y M kid. to be honest none of this is worth it
Answer:
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Explanation:
The lowering of the freezing point of a solvent is a colligative property ruled by the formula:
Where:
- ΔTf is the lowering of the freezing point
- Kf is the molal freezing constant of the solvent: 1.86 °C/m
- m is the molality of the solution
- i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.
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<u>a) molality, m</u>
- m = number of moles of solute/ kg of solvent
- number of moles of CaI₂ = mass in grams/ molar mass
- number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
- m = 0.0850667mol/1.25 kg = 0.068053m
<u>b) i</u>
- Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3
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<u>c) Freezing point lowering</u>
- ΔTf = 1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC
<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
I've always been told that it's eight. This is only because the atom has to have the saem protons (positively charged) neutrons and electrons (negatively charged). However this does not occur with all atoms. :)