Question

At 298 K , the osmotic pressure of a glucose solution is 10.50 atm . Calculate the freezing point of the solution. The density of the solution is $1.16 \mathrm{~g} / \mathrm{mL}$.

   At 298 K , the osmotic pressure of a glucose solution is 10.50 atm . Calculate the freezing point of the solution. The density of the solution is $1.16 \mathrm{~g} / \mathrm{mL}$.
 
Chemistry
Chemistry
Raymond Chang, Jason… 14th Edition
Chapter 12, Problem 136 ↓
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At 298 K , the osmotic pressure of a glucose solution is 10.50 atm . Calculate the freezing point of the solution. The density of the solution is $1.16 \mathrm{~g} / \mathrm{mL}$.
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At $298 \mathrm{K},$ the osmotic pressure of a glucose solution is 10.50 atm. Calculate the freezing point of the solution. The density of the solution is $1.16 \mathrm{g} / \mathrm{mL}$

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Transcript

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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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