The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.
The volume of the bottle = 4.7 L
Mass of ethanol = 0.33 g
Temperature (T1) = -11 oC = 273-11 = 262 K
P1 = 6.65 torr
Now we will calculate the mole by applying the ideal gas equation:-
PV = nRT
Or, n = PV/RT
Where P is the pressure
T is the temperature
R is the gas constant = 0.0821 L atm mol-1K-1
V is the volume
Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-
= 0.001913 mol C2H6
Conversion of the mole to gm
Molar mass of ethanol (M) = 46.07 g/mol
Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.
Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.
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<span>Your final answer would be C4H10O2, which equals 90amu</span>
<span>pv=nrt; Pressure and moles are constant.
p=nr(150k)/.5 L; Pressure initially
After temp change
pv=nrt; What is volume?
v=nr(350k)/p; p is constant so we can substitute from above
v=nr(350k)/(nr(150k)/.5 L))
v=350/150/.5 L
v=4.66 liters</span>
