Calculate the mass in grams of urea that must be added to150.g water to give a s

Question

Calculate the mass in grams of urea that must be added to150.g water to give a solution whose vapor pressure is 2.5 mm Hgless than that of pure water at 40 oC (vp H2Oat 40 oC =55.34 mm Hg). Show each step on how you solved this, include the correctunits used along the way with a clear explanation to recieve fullkharma points. Thanks for helping me understand. Calculate the mass in grams of urea that must be added to150.g water to give a solution whose vapor pressure is 2.5 mm Hgless than that of pure water at 40 oC (vp H2Oat 40 oC =55.34 mm Hg). Show each step on how you solved this, include the correctunits used along the way with a clear explanation to recieve fullkharma points. Thanks for helping me understand.

Explanation / Answer

We know that relative lowering of vapour pressure , (po-p ) / po = Mole fraction of solute                                                                                (po-p)/ po = n / ( n+N ) Where n = No . of moles of solute              = mass / Molar mass              = m /60                              Since Molar mass ofCN2H4O = 12 + 2*14 + 4*1 + 16 = 60g / mol N = No. of moles of solvent = mass / Molar mass                                            = 150 g / 18 g / mol                                            = 8.333 moles po = vapour pressure of pure solvent = 55.34 mm Hg p = vapour pressure of solution = 55.34-2.5 = 52.84 mmHg Plug the values we get 2.5 / 55.34 = (m/60 ) / ( (m/60)+ 8.333 ) On solving we get m = 2.128 *10^-3 g mass of Urea added = 2.128 *10^-3 g

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