At T = 373. K the vapor pressure of pure 1, 2 dibromoethane (E) is p_E* = 338. t

Question

At T = 373. K the vapor pressure of pure 1, 2 dibromoethane (E) is p_E* = 338. torr, and the vapor pressure of pure 1, 2 dibromopropane (P) is p_p* = 214. torr. a) What is the osmotic pressure for a 0.0114 mol/L of a nonvolatile solute dissolved in 12 dibromoethane at T = 373. K? Give your final answer in units of torr b) The total pressure above a liquid solution of 1, 2 dibromoethane and 1.2 dibromopropane at T = 373. K is P_total = 292. torr. Assuming these two liquids form an ideal solution, what is the value for X_E the mole fraction of 1, 2 dibromoethane in the solution?

Explanation / Answer

3. a) Osmotic pressure, = C X R X T

where C = Concentration = 0.0114 mol/L

R = Gas constant = 0.08206 L atm mol-1 K-1

T = Temperature = 373 K

= C X R X T = 0.0114 mol/L X 0.08206 L atm mol-1 K-1 X 373 K = 0.3489 atm = 265.16 torr

Osmotic pressure = 265.16 torr

b) PE* = 338 torr

PP* = 214 torr

Ptotal = 292 torr

Mole fraction of 1,2-dibromomethane, XE = Ptotal / PE*= 292 torr / 338 torr =0.864

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