When ethylene bromide (EtBr) and propylene bromide (PrBr) are mixed they form, t

Question

When ethylene bromide (EtBr) and propylene bromide (PrBr) are mixed they form, to a first approximation, an ideal solution. Consider a solution of EtBr and PrBr at T = 80.0 degree C. The vapor pressure above the solution is p = 139. torr. The vapor pressures of the pure liquids at this temperature are p*(EtBr) = 172. torr and p*(PrBr) = 127. torr. Find the mole fraction of EtBr in the solution and the partial pressure and mole fraction of EtBr in the vapor above the solution. The Henry's law constant for oxygen (O_2, M = 32.0 g/mol) in water, at T = 25.0 degree C, is K_H = 79.2 kPa/m^3 middot mol (recall 1 kPa = 1000 Pa). What is the concentration of dissolved oxygen in water at equilibrium with air (in units of mg/L) at T = 25.0 degree C? Assume an atmospheric pressure of 1.00 bar, and X(O_2) = 0.21 in air. Note that while the concentration is low, it is sufficient to support animal life in water.

Explanation / Answer

Let mole fraction of EtBr = x1

mole fraction of PrBr = x2 = 1-x1

P0 EtBr = 172 torr , P0PrBr = 127 torr ( vapor pressue of pure liquid )

& Ptotal = 127torr

According to raoults law

Ptotal = P01 x1 + P0 2 x1

Where P01 & P0 2 vapor pressure of pure liquids

X1 & X2 are mole fractions ( X1 + X2 = 1 )

put vapor pressure of EtBr & PrBr in raoults law to get molefraction

139 = 172X1 + 127( X2 = 1- X1 )

solving above equation

X1 ( EtBr) = 0.27 & X2 (PrBr)= 0.73

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