Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
As the gas cools it condenses and becomes a liquid its atoms also become smaller
<u>Analysing the Question:</u>
We are given a 250 mL solution of 0.5M K₂Cr₂O₇
Which means that we have:
0.5 Mole in 1L of the solution
0.125 moles in 250 mL of the solution <em>[dividing both the numbers by 4]</em>
<em />
<u>Mass of K₂Cr₂O₇ in the given solution:</u>
Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol
<em>we know that we have 0.125 moles in the 250 mL solution provided</em>
Mass = Number of moles * Molar mass
Mass = 0.125 * 194
Mass = 36.75 grams
