Benzene and toluene form ideal solutions. Consider a solution of benzene and toluene prepared at 25∘C. As suming that the mole f
ractions of benzene and toluene in the vapor phase are equal, calculate the composition of the solution. At 25∘C the vapor pressures of benzene and toluene are 95 and 28 torr, respectively.
2 answers:
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We can divide this problem into 4 steps:
Step 1: Calculate mass of one liter solution from the density
Mass of solution = 1000 ml soln x
=1005 g soln
Step 2: Calculate the mass of acetic acid
Mass of CH₃COOH = 1005 g soln x (5.2 g / 100 g soln) = 52.26 g acetic acid
Step 3: Calculate the moles of acetic acid:
52.26 g acetic x (1 mole acetic / 60 g acetic) = 0.871 mol
step 4: Calculate the molarity of acetic acid:
molarity = moles of acetic / liters of soln = 0.871 / 1 L = 0.871 mole / L
Step 1: Calculate mass of one liter solution from the density
Mass of solution = 1000 ml soln x
Step 2: Calculate the mass of acetic acid
Mass of CH₃COOH = 1005 g soln x (5.2 g / 100 g soln) = 52.26 g acetic acid
Step 3: Calculate the moles of acetic acid:
52.26 g acetic x (1 mole acetic / 60 g acetic) = 0.871 mol
step 4: Calculate the molarity of acetic acid:
molarity = moles of acetic / liters of soln = 0.871 / 1 L = 0.871 mole / L
Answer:
A) 0.831 M
B) 1.07 m
C) 26.2%
D) 0.0189
Explanation:
A) First we convert the mass of Pb(NO₃)₂ to moles:
- 275 g ÷ 331 g/mol = 0.831 mol Pb(NO₃)₂
Then we divide it by the total volume (1.00 L) to calculate the molarity:
- 0.831 mol / 1.00 L = 0.831 M
B) We convert the grams of water to kilograms:
- 775 g / 1000 = 0.775 kg
Then we divide the Pb(NO₃)₂ by the kilograms of water:
- 0.831 mol / 0.775 kg = 1.07 m
C) We divide the mass of Pb(NO₃)₂ by the total mass of the solution:
*100%
* 100% = 26.2%
D) We calculate the moles of water:
- 775 g H₂O ÷ 18g/mol = 43.1 mol H₂O
Then we divide the Pb(NO₃)₂ moles by the total number of moles:
= 0.0189