What is the vapor pressure at 75 degrees celcius of an aqueous solution prepared

Question

What is the vapor pressure at 75 degrees celcius of an aqueous solution prepared by the addition of 52.9 g of urea CO(NH2)2 to 178 g of water? The vapor pressure of pure water at 75 degrees celcius is 290 mmHg.

Explanation / Answer

FOLLOW THIS 2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O then.. via dimensional analysis.. 1 gal gas x (3780mL / gal) x (0.78g gas / mL) x (1 mole gas / 114.2g gas) x (16 moles CO2 / 2 moles gas) x (44.02g CO2 / mole CO2) x (1lb / 453.54g) = 20. lbs CO2 ********************** easiest way is to assume a known volume or mass and calculate from there. I'm going to assume we have 1000g of solution then.. Volume solution... 1000g x (1mL / 1.05g) x (1L / 1000mL) = 0.952 L mass E.G. 1000g solution x (40g E.G. / 100g solution) = 400.g moles E.G. 400g E.G. x (1 mole E.G. / 62.07g E.G) = 6.44 moles E.G. mass H2O 1000g solution - 400.g E.G. = 600. g H2O = 0.600kg H2O moles H2O 600.g H2O x (1 mole H2O / 18.02g H2O) = 33.3 and then... molarity = moles solute / L solution = 6.44 moles / 0.952 L = 6.76M molality = moles solute / kg solvent = 6.44 moles / 0.600kg = 10.7m mole fraction E.G. = moles E.G. / total moles = 6.44 moles / (6.44 moles + 33.3 moles) = 0.162 **************************** assuming Urea and sucrose are both non volatile, pH2O = pH2O* x mole fraction H2O where.. pH2O is the vapor pressure of the water over the solution pH2O* is the vapor pressure of pure water mole fraction of course is moles H2O / total moles moles H2O = 1kg x (1000g / 1kg) x (1 mole / 18.02g) = 55.5 moles sucrose = 85.5g x (1 mole / 342.3g) = 0.250 moles Urea = 75g x (1 mole / 60.06g) = 0.125 total moles = 57.0 mole fraction H2O = 55.5 / 57.0 = 0.974 so pH2O = 17.5torr x 0.974 = 17.0 torr

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