Question

A solution of 6.85 g of a carbohydrate in 100.0 g of water has a density of $1.024 \mathrm{~g} / \mathrm{mL}$ and an osmotic pressure of 4.61 atm at $20.0^{\circ} \mathrm{C}$. Calculate the molar mass of the carbohydrate.

   A solution of 6.85 g of a carbohydrate in 100.0 g of water has a density of $1.024 \mathrm{~g} / \mathrm{mL}$ and an osmotic pressure of 4.61 atm at $20.0^{\circ} \mathrm{C}$. Calculate the molar mass of the carbohydrate.
 
Chemistry
Chemistry
Raymond Chang, Jason… 14th Edition
Chapter 12, Problem 66 ↓
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A solution of 6.85 g of a carbohydrate in 100.0 g of water has a density of $1.024 \mathrm{~g} / \mathrm{mL}$ and an osmotic pressure of 4.61 atm at $20.0^{\circ} \mathrm{C}$. Calculate the molar mass of the carbohydrate.
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Transcript

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00:01 Hello everyone.
00:02 Thanks for joining me, ms.
00:03 Hallstrom, as we discuss osmotic pressure and use osmotic pressure to find molar mass.
00:09 And this problem will have just a little twist.
00:12 I've written down the three steps, and osmotic pressure is written as this pie right here equals molarity times gas constant times temperature in kelvin.
00:26 Malarity, gas constant, temperature and kelvin.
00:32 For this problem, we'll be solving for molarity.
00:38 Then we'll use that molarity right here.
00:42 Malarity equals moles per liter and we'll be solving for moles.
00:47 Finally, molar mass is calculated by dividing grams per mole.
00:52 The givens in this problem will be our osmotic pressure, our temperature, and mass of solvent.
01:03 The twist comes right here, and i'll put a little squiggly around that one.
01:09 I'll talk about that in a moment.
01:11 But first, we also know what the ideal gas constant is, because that's just a value we can look up if you don't have them memorized.
01:20 This particular problem, we are not going to be given volume.
01:25 We will be given mass masses of the solute and the solvent, so we can add those together to get the mass of the solution and we will be given density.
01:49 So from density and from the mass of the solution, we can find the volume.
01:55 Let's begin.
01:58 First, let's take a look at our data.
02:02 Our data for this problem, we are given mass of our solute, this is a carbohydrate, and we're given mass of solvent.
02:12 And these are 6 .85 grams and 100 .0 grams respectively.
02:28 And this is a carbohydrate and this is water.
02:33 So we're dealing with an aqueous solution today.
02:36 Okay.
02:38 We are also given a density of 1 .04 grams per milliliter.
02:54 And we are given our osmotic pressure of 461, 4 .61 atmospheres, and our temperature is 20 .0 degrees c, which is 293 .0 kelvin, and that should be 0 .0 degrees c.
03:25 Okay.
03:27 And the r we will be using, since we're dealing with atmospheres, 0 .08.
03:32 205 liter atmosphere over mole times k now let's go ahead and get started i'll start with the blue for step one what we're going to solve for in step one is malaria malaria is going to be equal to the osmotic pressure divided by r t let's go ahead and just solve that one we've got 4 .61 atmospheres divided by 0 .08205 latm over mole times k...
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