168 PROP0481: Detetmining the Molar Mass of a Volatile Liquid by the Dumas Metho

Question

168 PROP0481: Detetmining the Molar Mass of a Volatile Liquid by the Dumas Method 3. A student following the procedure of this experiment obtained the following data for an unknown volatile liquid: mass of flask, boiling stone, foil cap, 83.350 and unknown after cooling, g mass of flask, boiling stone, and foil cap,& water bath temperature, "C 82.657 95.0 30.09 barometric pressure, in. Hg volume of flask, mL 270 accepted molar mass of unknown, g mol" 86.2 (a) Calculate the mass of the unknown (b) Express the water bath temperature in K. (c) Express the barometric pressure in atmospheres.

Explanation / Answer

(a) Mass of flask + boiling stone + foil cap = 82.657gm

Mass of flask + boiling stone + foil cap +Mass of Unknown = 83.350gm

Mass of Unknown = ( Mass of flask + boiling stone + foil cap +Mass of Unknown ) - ( Mass of flask + boiling stone + foil cap ) = 83.350 - 82.657 = 0.693gm

(b) 0oC = (0 +273.16) Kelvin( K )

So, 95oC = ( 95 + 273.16 )K = 368.16K

(c) Barometic Pressure = 30.09in.Hg

1 inch-Hg = 0.0334 atmospheric pressure( atm )

So 30.09 inHg = 1.005006 atm

(d) Volume of Flask = 270mL

1 ml =1*10-3 L

270 mL = 0.27 L

(e) Density of gas is given by:

D = PM/RT

where P = Pressure of Gas = 1.005atm ; M = molecular mass of gas = 82.6 g mol-1 (given) ; R = constant = 0.082 L atm mol-1 K-1 ; T = Temprature = 368.16K

Therefore D = (1.005*82.6 )/( 0.082*368.16)

D = 2.75 g/L

(f) Molecular Mass = mRT/PV

where m = given mass of unknown = 0.693 gram ; V = volume =0.27 L ; other symbol have usual meaning

M = ( 0.693*0.082*368.16)/( 1.005*0.27 )

M = 77.1 g mol-1

(g) percent error = ( ( true value - observed value )/observed value )*100

% error = (82.6 - 77.1)*100/82.6

% error = 6.65%

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