Question

# 1.A) A student used the method MM=g*R*T/P*V for determine the molar masss of an unknown volatile...

1.A) A student used the method MM=g*R*T/P*V for determine the molar masss of an unknown volatile liquid. The barometric pressure in the lab was 746 mmHg, and the temperature of the boiling water was 98.5 degress C. The mass of the flask was 68.471 g, and the weight of the flask plus the condensed vapor was 687.959 g. The volume of the flask was 135.7 mL. What is the molar mass of the compound?

B) What would happen to the calculated MM in the results if the flask were heated too long? Explain.

C) Another student mistakenly measured a mass for the vapor that was larger than it should have been. Would the calculated molar mass be larger or smaller than it should be? Explain.

A.   MM=(g×R×T)/(P×V)

Ideal gas equation:    PV=nRT

Given data

P=746mm of Hg=1atm

T=98.50C=98.5+273.15=371.65 K

g= Mass of the gas= 687.959-68.471= 619.488g

R= Gas constant=0.08206 L.atm/mol .K

V=135.7×10-3=0.1357L

MM=(g×R×T)/(P×V)

= (619.488×0.08206×371.65)/(1×0.1357)

=139225 g/mol

B. Prolonged heating of the flask may not change the molar mass value as vaporization and condensation go simultaneously.

C. If the mass of the vapor was high(mistakenly measured), the calculated molar mass be larger than the actual value as molar mass is directly proportional to mass.

MM mass

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