What osmotic pressure in atmospheres would you expect for...

What osmotic pressure in atmospheres would you expect for each of the following solutions? (a) 5.00 $\mathrm{g}$ of $\mathrm{NaCl}$ in 350.0 $\mathrm{mL}$ of aqueous solution at $50^{\circ} \mathrm{C}$ (b) 6.33 $\mathrm{g}$ of sodium acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na},$ in 55.0 $\mathrm{mL}$ of aqueous solution at $10^{\circ} \mathrm{C}$

What osmotic pressure in atmospheres would you expect for each of the following solutions?
(a) 5.00 $\mathrm{g}$ of $\mathrm{NaCl}$ in 350.0 $\mathrm{mL}$ of aqueous solution at $50^{\circ} \mathrm{C}$
(b) 6.33 $\mathrm{g}$ of sodium acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na},$ in 55.0 $\mathrm{mL}$ of aqueous solution at $10^{\circ} \mathrm{C}$

Key Concepts

Molarity

Molarity is a measure of concentration defined by the number of moles of solute per liter of solution. It is essential in calculating colligative properties like osmotic pressure, where the concentration of dissolved particles directly influences the magnitude of the pressure.

Osmotic Pressure

Osmotic pressure is a colligative property representing the pressure required to stop the flow of solvent across a semipermeable membrane from a dilute solution to a more concentrated one. It is quantified by the formula ? = i M R T, highlighting how the pressure depends on the number of solute particles (via the van 't Hoff factor), the concentration of the solution (molarity), and the absolute temperature.

van't Hoff Factor

The van't Hoff factor (i) quantifies the number of particles a solute produces when dissolved. It is crucial in colligative property calculations because compounds that dissociate into multiple ions increase the effective solute particle count, thereby enhancing properties like osmotic pressure.

Colligative Properties

Colligative properties are physical characteristics of solutions that depend solely on the number of solute particles relative to solvent molecules, not on the specific chemical nature of the solute. Examples include osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure lowering.

Temperature Conversion

Temperature conversion is a vital concept in chemical calculations since many formulas, including those for colligative properties, require temperature input in Kelvin. Converting Celsius to Kelvin with the relation K = °C + 273.15 ensures proper use of the formula.

Transcript

00:01 All right, so here we're looking at an osmotic pressure question.

00:03 We're going to go ahead and write out our osmotic pressure equation.

00:08 I .m .r .t.

00:09 And then we'll go ahead and start with a and solve for what we know.

00:13 So first of all, on a, we're dealing with sodium chloride.

00:18 It's an ionic substance with two ions, which will generate two particles in solution.

00:23 So the van huff i factor is going to be two.

00:25 The malaria is equal to the moles of solute per liters of solution.

00:29 So we'll need to calculate that first.

00:32 So we have 5 grams of our solute in acl, and we'll need to convert that to moles using the molar mass of sodium chloride.

00:39 We'll take sodium 22 .99 and add that to chlorine 35 .45 to get a molar mass of 58 .44 per one mole.

00:50 By dividing 5 divided by 58 .44, we'll generate 0 .08, 56 moles of our nacl.

00:59 Now, for the liters portion of malaria, we'll be working with the 350 milliliters.

01:07 We know that it takes 1 ,000 milliliters to make one liter.

01:11 So if we divide by 1 ,000, that will tell us that we have 0 .350 liters of solution.

01:18 So for our malaria, we'll plug in 0 .0856 moles per 0 .3500 liters.

01:28 The r value is set at 0 .08, 2057, liters times atmospheres per kelvin times a mole.

01:36 That's the ideal gas constant.

01:38 And the temperature is given to us in celsius, so we'll need to convert that to kelvin by adding 273 .15.

01:46 So when you do that, 50 plus 273 .15, that generates a temperature of 323 .15 kelvin.

01:54 That will be the last factor that we plug in.

01:58 All right, so now let's multiply 2 times 0 .0856 divided by 0 .35 times 0 .057 times 323 .15...

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