This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the problems. Calculate the vapor pressure at 298K above an aerosol mixture consisting of 30\% w/w of aerosol propellant 114 (mol.wt = 170.9) with vapor pressure of 1.90 × 10^5 Nm^-2 and 70\% w/w of propellant 12 (mol.wt = 120.9) with vapor pressure of 5.85 × 10^5 Nm^-2. Assume ideal behavior.* Step 1: Assume a total mass of the mixture and calculate the mass of each propellant. Let's assume a total mass of 100 g for the mixture. Mass of propellant 114 (m_114) = 30\% of 100 g = 30 g Mass of propellant 12 (m_12) = 70\% of 100 g = 70 g Step 2: Calculate the moles of each propellant. Molar mass of propellant 114 (MW_114) = 170.9 g/mol Molar mass of propellant 12 (MW_12) = 120.9 g/mol Moles of propellant 114 (n_114): n_114 = 30 g170.9 g/mol = 0.17554 mol Moles of propellant 12 (n_12): n_12 = 70 g120.9 g/mol = 0.57900 mol Step 3: Calculate the mole fractions of each propellant in the liquid mixture. Total moles (n_total): n_total = n_114 + n_12 = 0.17554 mol + 0.57900 mol = 0.75454 mol Mole fraction of propellant 114 (X_114): X_114 = n_114n_total = 0.17554 mol0.75454 mol = 0.23265 Mole fraction of propellant 12 (X_12): X_12 = n_12n_total = 0.57900 mol0.75454 mol = 0.76735 Step 4: Calculate the partial vapor pressure of each propellant using Raoult's Law. Pure vapor pressure of propellant 114 (P_114^) = 1.90 × 10^5 Nm^-2 Pure vapor pressure of propellant 12 (P_12^) = 5.85 × 10^5 Nm^-2 Partial pressure of propellant 114 (P_114): P_114 = X_114 × P_114^ = 0.23265 × (1.90 × 10^5 Nm^-2) = 4.420 × 10^4 Nm^-2 Partial pressure of propellant 12 (P_12): P_12 = X_12 × P_12^ = 0.76735 × (5.85 × 10^5 Nm^-2) = 4.489 × 10^5 Nm^-2 Step 5: Calculate the total vapor pressure. Total vapor pressure (P_total) = P_114 + P_12 P_total = (4.420 × 10^4 Nm^-2) + (4.489 × 10^5 Nm^-2) P_total = (0.4420 × 10^5 Nm^-2) + (4.489 × 10^5 Nm^-2) P_total = 4.931 × 10^5 Nm^-2 The total vapor pressure is 4.93 × 10^5 Nm^-2. Calculate the vapor pressure (in Pa) at 20^ above an aerosol mixture consisting of 30\% w/w of HFA 134a (tetrafluoroethane, molecular weight 102) with a vapor pressure of 68.4 psig and 70\% w/w of HFA 227 (heptafluoropropane, molecular weight 170) with a vapor pressure of 56.0 psig. Assume ideal behaviour. NOTE: Psia = Psig + 14.7, 1 Psia = 6894.74 Pa* Step 1: Convert the pure vapor pressures from psig to psia, then to Pa. For HFA 134a: P_134a^ (psia) = 68.4 psig + 14.7 = 83.1 psia P_134a^ (Pa) = 83.1 psia × 6894.74 Pa/psia = 572953.554 Pa For HFA 227: P_227^ (psia) = 56.0 psig + 14.7 = 70.7 psia P_227^ (Pa) = 70.7 psia × 6894.74 Pa/psia = 487430.798 Pa Step 2: Assume a total mass of the mixture and calculate the mass of each HFA. Let's assume a total mass of 100 g for the mixture. Mass of HFA 134a (m_134a) = 30\% of 100 g = 30 g Mass of HFA 227 (m_227) = 70\% of 100 g = 70 g Step 3: Calculate the moles of each HFA. Molar mass of HFA 134a (MW_134a) = 102 g/mol Molar mass of HFA 227 (MW_227) = 170 g/mol Moles of HFA 134a (n_134a): n_134a = 30 g102 g/mol = 0.29412 mol Moles of HFA 227 (n_227): n_227 = 70 g170 g/mol = 0.41176 mol Step 4: Calculate the mole fractions of each HFA in the liquid mixture. Total moles (n_total): n_total = n_134a + n_227 = 0.29412 mol + 0.41176 mol = 0.70588 mol Mole fraction of HFA 134a (X_134a): X_134a = n_134an_total = 0.29412 mol0.70588 mol = 0.41666 Mole fraction of HFA 227 (X_227): X_227 = n_227n_total = 0.41176 mol0.70588 mol = 0.58334 Step 5: Calculate the partial vapor pressure of each HFA using Raoult's Law. Partial pressure of HFA 134a (P_134a): P_134a = X_134a × P_134a^ = 0.41666 × 572953.554 Pa = 238728.9 Pa Partial pressure of HFA 227 (P_227): P_227 = X_227 × P_227^ = 0.58334 × 487430.798 Pa = 284399.7 Pa Step 6: Calculate the total vapor pressure. Total vapor pressure (P_total) = P_134a + P_227 P_total = 238728.9 Pa + 284399.7 Pa = 523128.6 Pa Rounding to three significant figures: P_total = 5.23 × 10^5 Pa The total vapor pressure is 5.23 × 10^5 Pa. Drop the next question.