What is the vapor pressure of this solution?
Raoult's Law: Pi = Poi Xi Number of moles of sucrose: (177 g) / (342.3 g/mole) = 0.517 moles sucrose Number of moles of water: (624 g) / (18 g/mole) = 34.67 moles water Mole fraction of water: 34.67 / (34.67 + 0.517) = 0.985 Applying Raoult's Law: P = 31.8 mmHg * 0.985 = 31.3 mmHg.
Vapor pressure of water at 25 degree C is 23.76 mmHg.vapor pressure of solution containing 5.4 g dissolved in 90g water is 23.32 mmHg.?
Change in vapor pressure 23.76 - 23.32 = 0.44 fractional change = 0.44/23.76 = 0.0185 moles solute/total moles = mole fraction of solute moles water = 90 g x 1 mol/18 g = 5 moles x/x + 5 = 0.0185 0.0185x + 0.0926 = x x = 0.094 moles solute 5.4g/0.094 moles = 57.4 molar mass of solute
Calculate the vapor pressure of a solution?
moles glucose = 109 g/ 180.2 g/mol=0.605 mass water = 920 mL x 1 g/mL = 920 g moles water = 920 g/ 18.02 g/mol=51.1 mole fraction water = 51.1 / 51.1 + 0.605 =0.988 vapor pressure solution = 0.988 x 23.76 = 23.47 mm Hg
Determine the Vapor Pressure of a Solution that contains 66.6g of AlCl3?
Answer According to Raoult's law the vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. p = X solv + P0 Solv In this equation, Po is the vapour pressure of the pure solvent at a particular temperature. X solv is the mole fraction of the solvent., p is vap pr of soln. Vant Hoff factor has no contribution in vap pressure determination using Raoult's law http://en.wikipedia.org/wiki/Raoult's_la... In the given question, water is solvent, AlCl3 is a non volatile solute Molar mass of AlCl3 = 133.34 g / mol , moles AlCl3 = 66.6/133.34 = 0.4994 Moles of water = 250/18 = 13.88 moles [ given 66.6 g AlCl3 are available in 250 ml water ] Mole fraction of water = 13.88/[ 13.88+ 0.4944 ] = 0.9656 Vap pr of soln = 23.8*0.9656 = 22.98 torr [ vap pr of pure water * mole fraction of water [ AlCL3 has no contribution to vap pr as it a non volatile solute]
The vapor pressure of pure water at 25 °C is 23.8 torr.?
read up raoult's law of vapor pressure lowering for better understanding. There's a formula for this problem that you can use: Pa = P•a *Xa, where Pa is the vapor pressure above a solution containing glucose, P•a is the vapor pressure in a pure solvent and Xa is the mole fraction Vapor Pressure of H2O above a solultion = Vapor pressure of pure H2O * mole fraction of water to get the mole fraction: mole fraction of H2O = mole fraction of H2O/total number of moles in the soultion you'll have to convert the 13.4 g of glucose to moles 1 mole of glucose has 180 g and thus, 13.4 g will have .074 moles now convert the g of water to moles also 1 mole of H2O has 18 g. you're given 95 g of water, which has 5.27 moles mole fraction of H2O= moles of H2O/total moles= 5.27/5.34 = .98 plugging all this in the vp equation: Vapor Pressure of H2O above a solultion = Vapor pressure of pure H2O * mole fraction of water VP of H2O above solution = (23.8)*(.98) = 23.32 torr
Pls help! Calculate the vapor pressure of a solution of 34.0 g glycerol (C3H8O3) in 515.0 g water at 25°C.?
The vapor pressure of a solution is calculated with Raoult's law Psolution = Xsolvent * Psolvent where Xsolvent represents the mole fraction of the solvent and Psolvent is the vapor pressure of water in torr. To calculate the mole fraction of the solvent calculate the amount of moles of solute (C3H8O3 = 34.0g * (1 mole / 92.09382g) = 0.369189 mol. We're also going to need to convert H2O to moles to perform this calculation. 515.0g * (1 mole / 18.015g) = 28.58684 mol. (28.58684mol / 28.58684 - 0.369189) = 1.01308 * 23.76 torr = 24.1 torr For a more detailed explanation see cited source - it will step you through it really well.
As you would imagine, water vapor pressure changes with temperature. As you can see in the list below, water right before freezing has a very low vapor pressure of ~4.5 mmHg (or Torr). After 40C, the vapor pressure increases very rapidly, reaching 760 mmHg at 100C.Temperature,Water Vapor Pressure 0 C 4.58 mmHg 20 C 17.5 mmHg 40 C 55.3 mmHg 80 C 355.1 mmHg 100 C 760.0 mmHgAs far as boiling water is concerned, the vapor pressure is directly related to boiling point. By definition, water boils at the temperature where the atmospheric pressure equals the vapor pressure. Not surprisingly, you may notice that water at 100C would be boiling at the standard atmospheric pressure of 760 mmHg (or 1 atmosphere). If the atmospheric pressure were less, the boiling temperature of water would be lower (like say at Denver, CO). If you were to lower the pressure to near vacuum, you could basically make near-freezing water boil or ice sublimate! To understand this concept more intuitively, you have to imagine a microscopic bubble that cannot expand inside a liquid because the atmospheric pressure is forcing it to collapse. But if the pressure is the same as or greater than the atmosphere, then the bubble is able to expand and boil out of the liquid at any point inside the volume of water (rolling boil). Water, though, is always free to evaporate off the surface at any temperature which contributes to the vapor pressure.
Thanks for A2A.Xb = {Pa(pure solvent) -pa(solution)}/ Pa(pure solvent) ……….. (1)i.e. Relative lowering of vapor pressure is equal to the mole fraction of the solute.Here solute i.e. b = glucose, its weight (Wb)= 20g, Molecular mass (Mb)=180Solvent = water= a, weight (Wa)= 70g, Ma= 18Pa(pure solvent water) at 26 degree = 25.21, Pa(after dissolving solute) = ?Xb= no. of moles of solute/ no. of moles of (solute + sovent)…….. (2)no. of moles of solute= Wb/Mb= 20/180 = 0.112no. of moles of solvent = Wa/Ma = 70/18 = 3.89Putting in (2), Xb = 0.112/ (0.112 +3.89) = 0.0278Putting values of Xb, Pa(pure solvent), Pa(solution) in (1):0.0278 = {25.21 - Pa(solution)}/ 25.21Pa(solution) = 25.21- 0.0278*25.21 = 24.509 torr.hope it helps…… :)
Calculate the vapor pressure at 25 degrees Celcius of an aqueous solution that is 5.45 % NaCl by mass.?
This problem will involve finding the mole fraction of the solvent (in this case, water) and then using Raoult's law to find the vapor pressure as compared to normal water to determine the solution's vapor pressure. The first step to this problem is to find the mole fraction of water. First, find the number of moles of each solution component using a known quantity of solution. Using 100 grams will make using the mass percentage easy. Since the solution is 5.45% salt by weight, this means 100 grams of solution contains 5.45 grams of salt. Now, convert this to moles using the molar mass of NaCl, 58.44 grams per mole. This gives: 5.45/58.44 = .0933 mol of salt Now, since we know 5.45 grams of the solution of salt, the other 94.55 grams must be water. Divide the 94.55 grams of water by the molar mass of water, 18.02 grams per mole. 94.55/18.02 = 5.247 mole of water Now we know how many moles of each are in 100 grams of solution. Add these two quantities together to find the total number of moles: .0933 + 5.247 = 5.340 moles in solution Now we can determine the mole fraction of water in the solution. Divide the moles of water by the total moles: 5.247/5.340 = .9826 This is the mole fraction of water. Now we use Raoult's law, which states that the mole fraction of the solvent times the vapor pressure of the pure solvent equals the vapor pressure of the solution. The vapor pressure of pure water at 25 Celsius is 23.76 mmHg. Using Raoult's law, we get: 23.76 * .9826 = 23.34 mmHg The vapor pressure of this solution will be 23.34 mmHg Hope this explanation clears it up for you!
A solution contains 180 grams of glucose and 500 ml of water at 25 degrees C. What is the vapor pressure of the solution in torr?
Vapor pressure of solution = mole fraction of solvent X vapor pressure of pure solvent Moles glucose = 180 g / 180 g/mol = 1 mol glucose moles H2O = 500 mL X 1 g/mL X (1 mol / 18 g) = 27.8 mol H2O Mole fraction water = (27.8 mol / (1+27.8) = 0.9645 Vapor pressure solution = 0.9645 X 23.8 torr = 22.95 torr (I would round this to 23.0 torr)