Question

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C.

How many grams of testosterone, C₁₉H₂₈O₂, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 192.8 grams of diethyl ether to reduce the vapor pressure to 458.70 mm Hg ?

diethyl ether = CH₃CH₂OCH₂CH₃ = 74.12 g/mol.

Answer 1

P0-P/P0 = x

458.7/463.57   =x

0.989   = x

no of moles of diethyl ether = W/G.M.wt

                                              = 192.8/74.12 = 2.6moles

no of moles of solute ( testosterone) = W/288.4

P0-P/P0 = x ( mole fraction of solute)

458.7/463.57   = 2.6/2.6+z

0.989   =    2.6/2.6+z

0.989(2.6+z) = 2.6

   z   = 0.0289

no of moles of testosterone = W/G.M.Wt

                             0.0289   = W/288.4

                             W          = 0.0289*288.4   = 8.335g >>>> answer