1) If 1.71g of the vapor of a volatile liquid is able to fill a 390-mL flask at

Question

1) If 1.71g of the vapor of a volatile liquid is able to fill a 390-mL flask at 350C and 675mm Hg, what is the molecular weight of the liquid?

2) It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.05mL.

Assuming the density of water is 1.0g/mL, if 6 drops of water were left in the flask how many moles of water would this be?   

3) Using your answer for question 2, what volume(in mL) would this amount of water occupy when vaporized at 100oC and 1atm?

4) An Erlenmeyer flask that weighs 93.667g has a quantity of liquid vaporized in it at 96.4oC and a pressure of 1583.8 torr. The mass of the cooled flask is 94.345g. When the flask is filled with water, the mass of the flask and water is 268.92g.

(a). Determine the molecular weight of the unknown liquid

Explanation / Answer

1) If 1.71g of the vapor of a volatile liquid is able to fill a 390-mL flask at 350C and 675mm Hg, what is the molecular weight of the liquid

We will use ideal gas equation

PV = nRT

P = 675 mm Hg   V = 390mL = 0.39 L    R = 62.363 L mmHg K1 mol1 T = 35 + 273 = 308 K

So number of moles = 675 X 0.39 / 62.363 X 308 = 0.0137 moles

Moles = mass / molecular weight

0.0137 = 1.71 / molecular weight

Molecular weight = 1.71 / 0.0137 = 124.81 g / moles

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