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How many grams of sucrose '(C12Hz2On1) must be added to 563 g of water to give a solution with pressure 0.721 mmHg less than that o pure Water at 209C? (The av...

Question

How many grams of sucrose '(C12Hz2On1) must be added to 563 g of water to give a solution with pressure 0.721 mmHg less than that o pure Water at 209C? (The avapor mmHg vapor pressure Of water at 209C is 173

How many grams of sucrose '(C12Hz2On1) must be added to 563 g of water to give a solution with pressure 0.721 mmHg less than that o pure Water at 209C? (The avapor mmHg vapor pressure Of water at 209C is 173



Answers

How many grams of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ must be added to $552 \mathrm{g}$ of water to give a solution with a vapor pressure $2.0 \mathrm{mm} \mathrm{Hg}$ less than that of pure water at $20^{\circ} \mathrm{C} ?$ (The vapor pressure of water at $\left.20^{\circ} \mathrm{C} \text { is } 17.5 \mathrm{mmHg} .\right)$

So the mole fraction of water can be determined using the following expression of remote law P. Of H. 20 Is equal to the mole fraction H 20 Peanut H 20. Called the vapor pressure of the pure solving and vapor pressure of the solution. What we solve for XH 20. That's equal to 11955 millimeters of mercury, Divided by 1 4 9.44 mm of Mercury. What we get is not .7999. That is a mole fraction of water. And so we have the massive suit crossing grams. That can be determined using the mole fraction of water by using the expression for mole fraction of water, so that looks like follows more fraction of water is equal to the water moles divided by water moles, sucrose moles. While the moles of water, this is equal to 8.3333 moles. We know the mole fraction here, that's not .7999 moles. And again, of course, we know this value here. 8.333 moles. So we're looking to solve for the moles of Sucrose, Where we sold for this where X is equal to 2.08345 moles. And then what we do is multiply this by the weight 342 grounds per mole To get the mass 712.5399 g. Right?

So here we're looking to determine the mass of your area. Where firstly we need to look at the mole fraction of water, XH 20. And this is involved in a lower equations. We have p of water. Favorite pressure is equal to more fraction vapor pressure of the pure solvent. So the mole fraction of water, It's equal to 52.84 Miller m of market. We divide by 55.34 mm of mercury. We get 9.95482. And so we have, the mole fraction of water is equal to the moles of water, divided by the moles of water at the moles of your area, By the mold of water are 8.333 moles. So we've sold for this valley, we've sold for this valley. We have the mole fraction we've solved for that. And so all we're left with is our own note, the moles of your area. So we just rearrange soul for rex and for the most of your area, We should get 9.394- seven moles. We multiply that by 60.06 g per mole molecular weight to get 23.6 eight g.

A solution is made by dissolving 20.2 g to cross in 70.1 groundwater paper pressure. A pure water at 35 C is 42.2 million m uh market We have to calculate the lowering of paper pressure and the paper pressure of the solution at 35°C. Molecular weight obsolete grows. He's 342.30 ground part. So number of more options who grows Is 20.2 g. And divided by the molecular weight 3 42.30 Gramp are more So number of mobile apps across is 0.06. More. And number of moles. Mhm of water, 70.1 g water. We've headed by the molar mass of water which is 18.2 grandpa more And it is 3.89 more. So your .06 months across and 3.89 more water. Now small fraction of surplus. Yes, more obstacles divided by some of more obstacles plus malta water. So 0.06 divided by 3.95. So most fraction of sucrose is 0.015. Now, flooring of paper pressure. Delta P. Is be it zero that is the paper pressure up your solvent. And mole fraction of the salute here. So close And be a zero is 42.2. It is given. And Mole fraction of sucrose is 0.015. So that hyper pressure will be zero point 633 millimeter market. Okay. And now Delta B. Yes. The purpose of the pure solvent minus the per pressure of the solution P. A. We have to calculate the P. A. Is This 1- Delta P. And my purpose share of the pure solvent is 42.2 minus lowering of vapor pressure 0.633. So We put pressure of the solution 41.567. That is. We can write 41.6 millimeter Mark Gardy. There's that paper pressure up to solution at 35 C 41.6 millimeter marquetry and lowering of paper pressure. Delta P. Is 0.633 millimeter market.


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