(a) the vapour pressure of an ideal solution containing 32g of methanol and 138g
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Question
(a) the vapour pressure of an ideal solution containing 32g of methanol and 138g of ethanol is 0.3 atm at 27 C. When 46g more of ethanol is added to the same solution the vapour pressure of the solution increases by 0.013 atm at the same temperature. Calculate the vapour pressures of Ethanol and methanol in their pure state at 27 C. What is the partial vapour pressure of each of ethanol and methanol in the two solutions. (b) show from the following results, that acetic acid is associated in benzene solution and find its molar mass in benzene layer. Conc in water: 0.22, 0.36, 0.48, 0.67 conc in benzene: 0.027, 0.072, 0.128, 0.25Explanation / Answer
a) Both ethanol and methanol are the non volatilessubstances. Hence the vapor pressure of the solutionis - According to Raoult'slaw, P = P0 (CH3OH) *XCH3OH + P0(C2H5OH) *XC2H5OH where, P0 is the vapor pressure of the puresolvents given that the total vapor pressure = 0.3 atm Number of moles of methanol present = 32.0 g/32.0 g/mol= 1 mol Number of moles of ethanol = 138 g/ 46.0 g/mol = 3 mol total moles present initially = 4 mole Therefore, 0.3 atm = P0 (CH3OH) * (1/4)/+P0(C2H5OH) * (3/4) 0.3atm = 0.25 P0 (CH3OH) + P0(C2H5OH)0.75 -----------> 1 ) Step : 2 After adding the 46.0 g of ethanol to thesolution, the number of moles of the ethanol = 138 g+46/ 46.0g/mol =4 mol Therefore total moles of substances in the solution = 5mol The solution vapor pressure given = 0.3+0.013 atm = 0.313atm Therefore - 0.0039 atm = 1/5*P0 (CH3OH) +4/5*P0(C2H5OH) -------------> 2) By solving the 1) and 2) we get the vapor pressure ofpure ethanol = 0.365 atm vapor pressure of the methanol = 0.105 atm Therefore partial pressure of the methanol initially,P = P0*molefraction of methanol = 0.105 atm*0.25 = 0.02625 atm Partial pressure of ethanol = P0 *molefraction of ethanol = 0.365 * 0.75 = 0.27375 atm where, P0 is the vapor pressure of the puresolvents given that the total vapor pressure = 0.3 atm Number of moles of methanol present = 32.0 g/32.0 g/mol= 1 mol Number of moles of ethanol = 138 g/ 46.0 g/mol = 3 mol total moles present initially = 4 mole Therefore, 0.3 atm = P0 (CH3OH) * (1/4)/+P0(C2H5OH) * (3/4) 0.3atm = 0.25 P0 (CH3OH) + P0(C2H5OH)0.75 -----------> 1 ) Step : 2 After adding the 46.0 g of ethanol to thesolution, the number of moles of the ethanol = 138 g+46/ 46.0g/mol =4 mol Therefore total moles of substances in the solution = 5mol The solution vapor pressure given = 0.3+0.013 atm = 0.313atm Therefore - 0.0039 atm = 1/5*P0 (CH3OH) +4/5*P0(C2H5OH) -------------> 2) By solving the 1) and 2) we get the vapor pressure ofpure ethanol = 0.365 atm vapor pressure of the methanol = 0.105 atm Therefore partial pressure of the methanol initially,P = P0*molefraction of methanol = 0.105 atm*0.25 = 0.02625 atm Partial pressure of ethanol = P0 *molefraction of ethanol = 0.365 * 0.75 = 0.27375 atmRelated Questions
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