The vapor pressure of diethyl ether (ether) is 463.6 mm Hg at 25 °C. How many grams of aspirin , C9H8O4, a nonvolatile, nonelectrolyte (MW = 180.1 g/mol), must be added to 219.0 grams of diethyl ether to reduce the vapor pressure to 458.5 mm Hg ? diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.

526.3g of aspitin must be added

Explanation:

Using Raoult's law, the pressure of an ideal solution is equal to the vapour pressure of the pure solvent times mole fraction in the mixture.

The formula is:

As vapour pressure of the solution you want is 458.5mmHg and the vapour pressure of the pure solvent is 463.6mmHg:

458.5mmHg = X * 463.6mmHg

0.9890 = X.

That is:

0.9890 = Moles aspirin / Moles of diethyl ether.

Moles of diethyl ether are:

219.0g × (1mol / 74.12g) = 2.955 moles of diethyl ether.

That means moles of aspirin you need are:

0.9890 = Moles aspirin / 2.955 moles.

Moles aspirin = 2.922 moles.

As molar mass of aspirin is 180.1g/mol:

2.922 moles × ( 180.1g / mol) =