Gasoline is primarily a mixture of hydrocarbons and is sold with an octane ratin

Question

Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the properties of isooctane (C_8H_18), which has an enthalpy of vaporization of 35.8 kJ/mol and a boiling point of 98.2 degree C. Determine the vapor pressure of isooctane on a very hot summer day when the temperature is 38 degree 80.0 torr 36.7 torr 67.8 torr 47.9 torr 89.3 torr Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane. Gasoline usually contains an isomer of isooctane called tetramethylbutane (C_8H_18), which has an enthalpy of vaporization of 43.3 kJ/mol and a boiling point of 106.5 degree C. Determine the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38 degree C. 80.0 torr 36.7 torr 67.8 torr 47.9 torr 89.3 torr Calculate the molality of a solution containing 0.755 mol glucose and 1.75 kg of water. 0.875 m 0.583 m 0.431 m 580 m 0.850 m Calculate the molality of a solution containing 0.355 mol sodium hydrogen carbonate (baking soda) and 245 g of water. 1.45 m 1.12 Times 10^-2 m 2.11 m 1.12 Times 10^-3 m 0.0211 m What mass of a 0.66 m ammonium nitrate (80.04 g/mol) solution will contain 0.15 mol of solute? 227 g 239 g 440 g 493 g 4453 g How many moles of solute are there in a 0.155 m glucose solution prepared with 50.0 kg of water? 15.5 mol 15.0 mol 0.155 mol 31.0 mol 7.75 mol What is the molarity of a sucrose (C_12H_22O_11) solution that produces an osmotic pressure of 2.65 atm at 25 degree C? 0.0349 M 0.127 M 0.0127 M 0.108 M 0.398 M

Explanation / Answer

28. Using,

ln(P2/P1) = dHv/R[1/T1 - 1/T2]

P1 is vapor pressure at 38 oC

Feding the values,

ln(760/P1) = 35800/8.314[1/311 - 1.371.2)

P1 = 80 torr

a. 80.0 torr

29. Again using the above formula,

ln(760/P1) = 43300/8.314[1/311 - 1/379.5)

P1 = 36.7 torr

b. 36.7 torr

30. Molaltity = moles of solute/kg of solvent

So, molarity of solution = 0.755 mol/1.75 kg = 0.431 m

c. 0.431 m

31. molarity of solution = 0.355 mols/0.245 kg = 1.45 m

a. 1.45 m

32. mass of NH4NO3 = 0.15 mol/0.66 m = 0.227 kg = 227 g

a. 227 g

33. moles of solute = 0.155 m x 50 kg = 7.75 mol

e. 7.75 mol

34. osmotic pressure = molarity x gas constant x temperature

molarity of solution = 2.54/0.082 x 298 = 0.108 M

d. 0.108 M

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