please show step by step solutions. Thank you.
please show step by step solutions. Thank you. 13. The vapor pressure of benzene (C6H6) is...
5) The vapor pressure of liquid benzene is 20,170 Pa at 298.15 K, and AHvap = 30.72 kJ/mol at 1.00 atm. Calculate the normal boiling point (boiling point at 1.00 atm) of benzene.
6.34. The vapor pressure of benzene at 40.0°C is 0.241 atm. If the enthalpy of vaporization of C. Ha is 33.9 kJ/mol, estimate the normal boiling point of benzene.
The heat of vaporization of benzene, C6H6 is 30.7 kJ/mol at its boiling point of 80.1 C. How much energy in the form of heat is required to vaporize 132 g benzene at its boiling point? a) 0.302 kJ b) 51.9 kJ c) 24.2 kJ d) 40.1 kJ c) 4.05 x 10^3 kJ
At 25C, the vapor pressure of pure benzene is 0.1252 atm. When 6.50 g of a sold sample of naphthalene,C10H8(128.17 g/mol), is dissolved in 76.0 g of pure benzene, C6H6, calculate the vapor pressure of benzene above the solution oy looL 2. Raoult's Law. At 25 °C, the vapor pressure of pure benzene is 0.1252 atm. When 6.50 g of a solid sample of naphthalene, C10Hg (128.17 g/mol), is dissolved in 76.0 g of pure benzene, C6H6 (78.0 g/mol), calculate...
Benzene, C6H6, and octane, C8H14, form an ideal solution. At 60°C the vapor pressure of pure benzene is 0.507 atm, and the vapor pressure of pure octane is 0.103 atm. A solution is composed of 3.53 g of benzene, and 40.2 g of octane. What is the mole fraction of benzene in the vapor phase above the liquid? In order to receive full credit, your work should clearly show the following: a) the calculation of the partial pressure of benzene...
Can you show the work for these problems? Ethanol has an enthalpy of vaporization of 42.3 kJ/mol. The compound has a vapor pressure of 1.00 atm at 78.3 degree C. At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K middot mol) -83.8 degree C -24.4 degree C 62.6 degree C 73.0 degree C 78.0 degree C A liquid has an enthalpy of vaporization of 30.4 kJ/mol. At 269 K is has a vapor pressure...
The vapor pressure of Y is 4.590 atm at 330K and 9.180 atm at 380K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). AG330º = kJ/mol AG380º = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values ASO = J/(mol-K) AH = kJ/mol What is the normal boiling point of Y in degrees Celcius? t = Ос What...
The vapor pressure of Y is 2.320 atm at 420K and 5.336 atm at 478K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(u) - Y(9). kJ/mol AG420º = AG478° - kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values 3/(mol-K) AS- AH kJ/mol What is the normal boiling point of Y in degrees Celclus? oc What is the vapor pressure...
Given the heat of vaporization of benzene as 30.8 kJ/mol and the vapor pressure = 92 mm Hg at 25°C. Calculate the vapor pressure of benzene at 75°C.
2. Raoult's Law. At 25 °C, the vapor pressure of pure benzene is 0.1252 atm. When 6.50 g of a solid sample of naphthalene, C10Hg (128.17 g/mol), is dissolved in 76.0 g of pure benzene, C6H6 (78.0 g/mol), calculate the vapor pressure of benzene above the solution. [0.119 atm]