At 298 K, the osmotic pressure of a glucose solution is...

Key Concepts

Concentration Conversion

Concentration conversion involves converting between different measures of solution concentration such as molarity (moles of solute per liter of solution) and molality (moles of solute per kilogram of solvent). This conversion is important when applying formulas like the freezing point depression equation, which requires the molality of the solution, especially when the density of the solution is provided and such a conversion is necessary.

Freezing Point Depression

Freezing point depression is a specific colligative property where the addition of a solute to a solvent lowers the freezing point of the solvent. It is quantified by the equation ?Tf = iKfm, where ?Tf is the temperature change, i is the van't Hoff factor (which represents the number of particles the solute dissociates into; for non-electrolytes like glucose, i = 1), Kf is the freezing point depression constant, and m is the molality of the solution. This concept is vital in connecting solute concentration with changes in phase transition temperatures.

Colligative Properties

Colligative properties are properties of solutions that depend on the number of solute particles relative to the solvent molecules, not on the identity of the solute. These properties include freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure, all of which are essential for understanding how the presence of a solute alters the physical behavior of a solvent.

Osmotic Pressure

Osmotic pressure is a colligative property that reflects how the concentration of solute particles in a solution creates a pressure difference across a semipermeable membrane. It is given by the formula ? = cRT, where ? is the osmotic pressure, c is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin. This concept is pivotal when relating solute concentration to physical properties of solutions.

Transcript

00:01 Hello everyone and thanks for joining me, ms.

00:03 Halsstrom, as we calculate the freezing point for a solution.

00:07 This is a pretty challenging problem, but i'm going to give you a little plan here.

00:13 For this problem, we're going to have several givens.

00:16 I'll outline those in a moment, but we are going to use raoults law to solve for malaria.

00:33 We will be using what we know about density to solve for mass, of our solution, we'll be given all the data we need to do these things, we will find moles of solute, and after moles of solute, we will use that to find our delta t in our freezing point, and since it's water, it'll be very easy to tell us what the freezing point is.

01:12 Okay, i'm going to go to the next page to put our data down, and for this problem, we know the following information.

01:22 We are at 298 kelvin and we have an osmotic pressure, make that a little nicer, our temperature is 298k, our osmotic pressure is 330 or excuse me 10 .50 atmospheres and our density of the solution in question is 1 .1 .1 .1 .000.

01:59 16 grams per milliliter.

02:06 Okay, as you recall, step one.

02:11 Step one, we're going to use roald's law where the osmotic pressure, multiplied by malarity, times the ideal gas constant, times t, will be applied.

02:26 And we're going to solve for m.

02:30 Rearranging for m and substituting my values, i get 10 .50 atmosphere.

02:38 Divided by, we're going to use the ideal gas constant of 0 .08205 latm over k -m because the units for pressure match, and our temperature is 298 kelvin.

02:57 Doing the math for this, i get a malarity of 0 .494.

03:08 That value will be used again.

03:13 Okay, that is step one.

03:16 For step two, we are going to use density.

03:27 We're going to use our knowledge of density and the density that we're given to figure out the mass of our solution.

03:40 So now we know that the density is equal to mass of substance divided by volume of substance.

03:57 Okay, so our density of solution is 1 .16 grams per milliliter and we're dealing with a one a solution where we've used molarity so that i know i have one liter i have one liter of solution because our value right here molarity is moles per liter one liter of solution equals 1 ,000 milliliters of solution we're going to multiply that by 1 .16 grams per milliliter.

04:39 And you don't even need a calculator for this one.

04:42 I get 1 .1 ,160 grams, which i'm going to right now put as 1 .16 kilograms...

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