Question #02e7c
1 Answer
Answer:
Explanation:
I'll assume these are the measurements when the liquid and vapor are at dynamic equilibrium.
To solve this problem, we can first use the idealgas equation:
to calculate the number of moles of the gas.
We're trying to find the number of moles,
#n = (PV)/(RT)
We know

#P = 755cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = 0.993# #"atm"# 
#V = 498cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.498# #"L"# 
#R = 0.082057("L"·"atm")/("mol"·"K")# 
#T = 100^"o""C" + 273 = 373# #"K"#
Plugging these into the equation, we have
We know that the mass of the vapor is
This is the mass of one mole of the vapor. The question asks for its molecular weight, which is the same as its atomic mass. The mass of one mole of a substance is the mass in
Therefore,
The molecular weight of the gas is thus