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Calculate the mass in grams of urea that must be added to 150.g water to give a

ID: 691577 • Letter: C

Question

Calculate the mass in grams of urea that must be added to 150.g water to give a solution whose vapor pressure is 2.5mm Hg lessthan that of pure water at 40oC (vp H2O at 40oC=55.34mm Hg) Show each step, clear explanation to recieve life saverpoints, Thanks for the help Calculate the mass in grams of urea that must be added to 150.g water to give a solution whose vapor pressure is 2.5mm Hg lessthan that of pure water at 40oC (vp H2O at 40oC=55.34mm Hg) Show each step, clear explanation to recieve life saverpoints, Thanks for the help

Explanation / Answer

According Raoult's law relative Lowering of vapourpressure ( po-p) / po = mole fraction of the solute                                                                                           ( po - p ) / po =n / ( n+ N ) where po - p = lowering of vappour pressure = 2.5 mm Hg                 po = vapour pressure of pure water = 55.34 mm Hg n = No . of moles of urea = mass / Molar mass                                       = m / 60 g / mol N = No . of moles of water = mass / Molar mass                                           = 150 g / 18 g/mol                                           = 8.334 moles n+N = ( m/ 60 ) + 8.334 Plug the values we get ( 2.5 / 55.34 ) = ( m/60 )/ [ ( m/ 60 ) + 8.334 ]                                      0.045 = (m/60 ) / [ m +( 60*8.334 ) / ( 60 )]                                      0.045 = m / ( m + 500.04)                                   0.045 * ( m + 500.04 ) = m                                 0.045 m + 22.50 = m                                 m-0.045 m = 22.5                                      0.955 m = 22.5                                                m = 22.5 / 0.955                                                     =23.562 g On solving we get mass of urea added m = 23.562 g Edit : calculation error.
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