2. Would the molar mass of a volatile liquid, calculated using the procedure in

Question

2. Would the molar mass of a volatile liquid, calculated using the procedure in this experiment, be incorrectly high, incorrectly low, or unaffected by the following procedural changes? (a) You did not completely vaporize the liquid when you heated it. (b) The foil cap got wet while you were cooling the flask and its contents with running water. (c) You added the boiling stone to the flask after you had already determined the mass of the empty flask and foil cap. (d) You forgot to measure the volume of the flask, so you used the volume printed on the flask for your calculations. (e) Your unknown liquid had a boiling point of 102.3 Celsius.

Explanation / Answer

In Dumas method for determination of molar mass, the unknown volatile liquid is taken in a flask and the flask is sealed and kept in a boiling water bath. The unknown liquid gradually gets converted to vapor. The pressure of the vapor is used to calculate the moles of unknown substance (n) by applying the ideal gas equation PV = nRT.

(a) When the liquid was not completely vaporized then pressure value was less than actual. As a result, the value of n would be less. Molar mass is calculated by dividing the mass of liquid by number of moles. For this reason, here the molar mass value would be incorrectly high.

(b) When the foil cap got wet it retained some water and this increased the value of vapor pressure. As a result, value of n would increase and consequently the molar mass value would be incorrectly low.

(c) When boiling stone was added after determining the mass of empty flask and foil cap then its mass was added to the mass of the unknown liquid. As a result, the molar mass value was incorrectly high.

(d) Volume measured from the volume printed on the flask was correct and the value of molar mass was unaffected.

(e) If the unknown liquid had a boiling point higher than water then they would never boil because the boiling water would prevent the water from going over 100°C. As a result, pressure value would be low and the molar mass value would be incorrectly high because molar mass = mass / number of moles.

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