(1)What effect does the presence of a nonvolatile solutehave upon (a) the vapour
ID: 676463 • Letter: #
Question
(1)What effect does the presence of a nonvolatile solutehave upon (a) the vapour pressure of a solution, (b) the freezingpoint and (c) the boiling point? (2) what is the molality of a solution that contains 3.0g urea( molecular weight = 60 amu) in 100g of benzene. (3) calculate the freezing point of a solution containing2.50g of benzene in 120g of chloroform. (4) A solution contaning 2.00g of an unknown substance in25.0g of naphthalene was found to freeze at 75.4 degrees celsious.what is the molar mass of the unknown substance? (9) How many grams of NaNO3 would you add to 250g of water inorder to prepare a solution that is 0.500 molal in NaNO3? (1)What effect does the presence of a nonvolatile solutehave upon (a) the vapour pressure of a solution, (b) the freezingpoint and (c) the boiling point? (2) what is the molality of a solution that contains 3.0g urea( molecular weight = 60 amu) in 100g of benzene. (3) calculate the freezing point of a solution containing2.50g of benzene in 120g of chloroform. (4) A solution contaning 2.00g of an unknown substance in25.0g of naphthalene was found to freeze at 75.4 degrees celsious.what is the molar mass of the unknown substance? (9) How many grams of NaNO3 would you add to 250g of water inorder to prepare a solution that is 0.500 molal in NaNO3?Explanation / Answer
a ) i) Vapour pressure of the solution is lowered. ii ) Freezing point is depressed iii) Boiling point is increased. b) molality = Number of mols / Mass of solvent inkg Number of mols = 3.0 g / 60.07 g / mol = 0.0499mols molality = 0.0499 mols /0.1000 kg = 0.499 m c ) Tf = Kf m m = 2.50 g / 78.11g/mol x 0.120 kg = 0.266 m Kf = 5.12 0 C /m Tf = 5.12 0 C / m x0.266 m = 1.370 C Tf = T pure -Tsoln Tsoln = T pure-Tf =5.5 0 C - 1.37 0 C =4.13 0 C d) Tf = T pure - Tsoln = 80.26 °C -75.4 °C = 4.86 °C Tf = Kf x m m = Tf / Kf = 4.86 °C/ 6.80 °C/m =0.714 m molar mass = 2 g / 0.714 m x 0.025 kg =112.04 g /mol d) Tf = T pure - Tsoln = 80.26 °C -75.4 °C = 4.86 °C Tf = Kf x m m = Tf / Kf = 4.86 °C/ 6.80 °C/m =0.714 m molar mass = 2 g / 0.714 m x 0.025 kg =112.04 g /mol 9) 0.500 molal = mass in g / molar mass in g / molx Mass of solvent in kg =mass in g / 84.9948 g/mol x 0.25 kg mass ing = 0.500 molal x 84.9948 g/molx 0.25 kg = 10.6244g`
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