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owv2] Online teaching com/ilm/takeAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionLocator-assignment-take; Use the References to access important values if needed for this question The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C How many grams of aspirin, C Hs04 a nonvolatile, nonelectrolyie M to 456.06 mm Hg? ·is i g mol must be added to lSO3 grans of diethyl ether to reduce the vapor pressure diethyl ether = CH3CH2OCH,CH,-74.12 g mol. g aspirin Submit Answer Try Another Version 1 item attempt remaining

Solution

According to Ravoult\'s law

P_solution = X_solvent P°_solvent

   P_solution = 456.06mmHg

P°_solvent = 463.57mmHg

Therefore,

X_solvent = 456.06mmHg/463.57mmHg =0.9838

Therefore,

mole fraction of diethyl ether must be 0.9838

mole fraction of diethyl ether = no of moles of diethyl eather/total no of moles

No of mole of diethy ether = 180.3g/74.12g/mol = 2.4325

0.9838= 2.4325/Total no of moles

Total no of moles = 2.4325/0.9838 = 2.4726

No of moles of Aspirin = 2.4726 - 2.4325 = 0.0401

Molar mass of Aspirin = 180.1g/mol

Mass of aspirin should be added = 180.1g/mol × 0.0401mol = 7.2220g