Arrange the following solutions in order by their decreasing...

Arrange the following solutions in order by their decreasing freezing points: $0.1 \mathrm{m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.1 \mathrm{m} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ $0.01 m \mathrm{CO}_{2}, 0.15 \mathrm{m} \mathrm{NaCl},$ and $0.2 \mathrm{m} \mathrm{CaCl}_{2}$

Key Concepts

Electrolytes vs. Non-Electrolytes

This concept distinguishes between solutes that dissociate into ions (electrolytes) and those that do not (non-electrolytes). Electrolytes produce more particles per formula unit when dissolved, significantly influencing colligative properties such as freezing point depression, relative to non-electrolytes of the same molal concentration.

Van’t Hoff Factor

The van’t Hoff factor is a measure of the number of particles into which a solute dissociates in solution. It is crucial because colligative properties depend on the total number of solute particles present. For example, electrolytes that dissociate into multiple ions have a higher effective concentration of particles than non-electrolytes, which affects freezing point depression.

Freezing Point Depression

Freezing point depression is a colligative property where the addition of solute particles to a solvent lowers its freezing temperature. The extent of the temperature decrease is directly proportional to the concentration of solute particles dissolved, meaning more particles lead to a larger depression of the freezing point.

Colligative Properties

Colligative properties refer to the properties of solutions that depend primarily on the number of solute particles in a given amount of solvent, rather than the nature or identity of the particles. This principle is essential for understanding phenomena like freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

Transcript

00:02 So we are given all of these 0 .1 molar, na3po4, 0 .1 molar, c2h5oh, ethanol, 0 .01 molar, co2, 0 .15 molar, nacl, and 0 .2, molar c a cl 2 and we have to arrange in decreasing order of freezing point okay so we'll start with a general formula which is delta t equals to i k f into b b is molality k f is a constant and i is wanthoff factor.

01:02 So all of these are going to be a constant, right? so let's start.

01:09 So they will not vary.

01:10 So what varies is actually this value.

01:13 So we'll start with the first one, na3, po4.

01:20 So molality of na3 po4 is equal to 0 .1 mole per kg, right? so if you look at i n.

01:36 3 p .04 aqueous will dissociate as sodium ions plus p .04, 3 negative ions.

01:52 Therefore i is equal to 4 because there are 3 sodium and 1 potassium ion 1 p .ofor ions, right? so delta t n2, so4 is equal to 0 .1 mole kg into 4 into kf kf is going to be a constant throughout so this is 0 .4 mole kg kf okay so this is the first one moving on to the second one which is 0 .1 molar c2h5 ohh so b c2h5 oh is equal to 0 .1 mol per kg now the dissociation part is that i will be equal to one as it will not dissociate therefore delta t c2h5 oh is equal to 0 .1 mole kg into 1 into k f so this is just 0 .1 mole kg kf okay moving on to the third one 0 .01 molar co2 the co2 is 0 .01 more per kg and co2 is also not dissociating so i co2 is equal to 1 delta t co2 is equal to 0 .01 more or kg into kf okay moving on to the fourth one 0 .15 molar nacl so in this case b nacl or molality is 0 .15 moles per kg and nacl aquas will dissociate to form any positive aquas plus cle negative aquas therefore i nacl is equal to 2 because there are total of 2 ions so this gives delta d nacl equals to 0 .15 mole kg into 2 into k f so this is 0 .3 mole per kg kf moving on to the last one 0 .2 molar cacl 2 so, molality is 0 .2 moles per kg...

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