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1.20 g of an unknown liquid is placed into a 2.3 L container under vacuum. (No...

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Chemistry

1.20 g of an unknown liquid is placed into a 2.3 L containerunder vacuum. (No gases present.) The liquid completely evaporatesat 25.03°C and becomes a gas with a pressure of 325.3 torr. What isthe molar mass of the the unknown substance?

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3.9 Ratings (767 Votes)

volume = V= 2.3L

Pressure = p= 325.3 torr = 0.428 atm

mass of gas = w = 1.20 grams

Temperature = T = 25.03C = 298.03 K

Gas constant = R= 0.0821 L-atm/k/mole

Ideal gas equation =        PV=nRT   , but n= mass/molar mass =    w/M

                                       PV= w/M*RT

                             M = w xR xT/PV

                                 = 1.20 x 0.0821 x 298.03/0.428x2.3

                                    = 29.36/0.9844 = 29.83 grams

Molar mass of unknown gas = 29.83 grams.
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