Question

B. The following questions from the last experiment, PROP 481 a. Calculate the density of gaseous diethyl ether at 889 Cand 0.966 atm. The molar mass of diethyl ether is 74.12g/mol b. A student removes the flask from the hot water bath before it was completely converted into its vapor? Will the determined molar mass be higher, or unaffected? i) if you are given liquid that has a higher boling point than water, what modification you will adopt to find the molar mass? i) A student left the flask with the liquid in the hot water bath more time after its complete evaporation, what is its effect on the results of the experiment? Will the molar mass be higher or lower than the actual value?

Answer 1

1) Using equation, d=P*M/RT

R=0.0821Latm/Kmol

P=0.966 atm

T=88.9+273=361.9K

M=74.12 g/mol

d=(0.966 atm)(74.12 g/mol)/(0.0821Latm/Kmol)(361.9K)=2.410g/L=2.410g/1000ml=0.00241 g/ml

Explanation:

Molar mass determination of volatile liquid:

To determine the molar mass (M)of a volatile liquid,the liquid is evaporated in a flask of measured volume(V) being submerged in boiling water,so that the temperature(T) of water equals that of gas in the flask.After the liquid condenses,the mass of the flask+liquid is measured and substracted from the mass of flask to calculate the mass of liquid(m).Barometric pressure (P)is also noted and all the values plugged in the ideal gas equation,PV=nRT=m/M RT

to calculate density: d=mass/volume=m/V

PV=m/M RT

or,P=(m/V)(RT/M)=d*(RT/M)

d=P*M/RT