Calculate the osmotic pressure of each of the following aqueous solutions at 20^∘ C: a. 2.39 M m

Calculate the osmotic pressure of each of the following...

Key Concepts

Temperature Conversion

Accurate calculation of osmotic pressure requires the use of absolute temperature in Kelvin rather than Celsius. Converting Celsius to Kelvin (by adding 273.15) ensures that the gas constant R is applied correctly in the osmotic pressure equation, maintaining consistency in scientific calculations.

Solution Concentration

Solution concentration, often expressed in molarity (moles per liter), is critical in determining the number of solute particles present in a solution. Since osmotic pressure is directly related to the concentration of solute particles, understanding how to convert and calculate concentration from various units is key for solving related problems.

van’t Hoff Factor

The van’t Hoff factor (i) defines the effective number of particles produced in solution from a dissolved compound. For non-electrolytes that do not dissociate, i is typically 1, whereas for electrolytes that dissociate into multiple ions, i exceeds 1. This factor is essential in the accurate calculation of osmotic pressure and other colligative properties.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of solute particles present rather than their specific identity. These properties, which include osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure lowering, are valuable for determining aspects like molecular weights and help in understanding solution behavior upon dilution.

Osmotic Pressure

Osmotic pressure is a colligative property that measures the pressure needed to stop the net flow of solvent molecules through a semipermeable membrane from a less concentrated solution to a more concentrated one. It is quantitatively described by the equation ? = i M R T, where ? is the osmotic pressure, i is the van’t Hoff factor, M is the molar concentration of the solute, R is the gas constant, and T is the absolute temperature in Kelvin.

Transcript

00:01 So we have these various solutions, and we want to calculate their osmotic pressure at 20 degrees c.

00:11 And the equation for osmotic pressure, if you're working with a property that is defined by a mathematical equation, i'll always begin by writing that equation first.

00:25 Solve it for the unknown.

00:27 Here, the unknown is osmotic pressure, so it's already solved for the osmotic.

00:31 Emotic pressure, and we have 2 .39 molar methanol.

00:37 Now, methanol is a non -electrolite.

00:43 So when one mole of it dissolves, we get one mole of methanol, aq.

00:48 So there is no factor to take into account there.

00:52 So it's just really a matter of multiplying the molarity, which i've written as moles per liter, because units have to work.

01:01 Anytime you use, are the ideal gas constant.

01:05 You've got to be sure that you're using the same units that are in it.

01:08 And i am using moles and liters and kelvin's.

01:13 And so i will get units of pressure for my answer.

01:19 And multiplying that out, that is incredible.

01:41 57 .5 atm.

01:46 That is a tremendous amount of pressure.

01:50 That's one of the reasons why osmotic pressure is so useful, is it doesn't take a very concentrated solution to give quite a bit of pressure, and one can detect that then.

02:00 So now, i'm not going to do the b part.

02:04 I'm going to tell you, though, that that 9 .45 millimolar magnesium chloride will give you 9 .45 millimolar magnesium ion plus, 18 .9 millimolar chloride ion because you get two moles of chloride for each mole of the compound that dissolves and one mole of magnesium.

02:48 And so when you go to plug in the molarity into the osmotic pressure equation, you need not to plug in 9 .45, but let's see, that's 27, 28 .35.

03:04 And that's milly, so times 10 to the minus 3, molar.

03:10 And other than that, you will work the b part exactly the same as the a.

03:15 Now, in the c part, we're given the volume of glycerol that was mixed with enough water to make 250 milliliters of an aqueous solution.

03:27 And so we've got to use the density formula, multiplying both sides by v.

03:33 Gives me the dv equals mass, take the given density, and multiply it by the 40 milliliters...

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