Answer:
hope that helps
How many moles of aluminum oxide (Al2O3) are in a sample with a mass of 204.0 g? The molar mass of aluminum oxide (Al2O3) is 102
1 answer:
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Answer : The vapor pressure of hexane = 140.5055 torr
Solution : Given,
Concentration of Naphthalene by mass = 10 %
Vapor pressure of pure Hexane = 151 torr
From the periodic table, the molar masses of Naphthalene and Hexane are:
Molar mass of Naphthalene = 128 g/mole
Molar mass of Hexane = 86 g/mole
From the Raoult's law, the vapor pressure of the component is equal to the vapor pressure of the pure component multiplied by its mole fraction.
Formula used :
..............(1)
Where,
= vapor pressure of component i
= vapor pressure of pure component i
= mole fraction of of component i in the solution
For this problem, few steps are involved:
step 1 : concentration of 10% naphthalene by mass means 10 g of naphthalene present in 100 g of solution.
This means,
The mass of naphthalene = 10 g
The mass of hexane = 100 - 10 = 90 g
step 2 : find the number of moles of naphthalene and hexane.
Formula used :
Number of moles =
Moles of Naphthalene = = 0.0781 moles
Moles of Hexane = = 1.0465 moles
Total moles = Moles of Naphthalene + Moles of Hexane
= 0.0781 + 1.0465 = 1.1246 moles
step 3 : Now, calculate mole fraction of Hexane
Mole fraction of Hexane = =
= 0.9305
step 4 : Now, find the vapor pressure of Hexane by using above formula (1).
Now put all the values in this formula, we get
Vapor pressure of Hexane = Vapor pressure of pure Hexane × Mole fraction of Hexane
= 151 torr × 0.9305 = 140.5055 torr
Answer:
The concentration of H₂C₂O₄ is 7,1800x10⁻³ M
Explanation:
The reaction of the titration is:
2 KMnO₄ + H₂C₂O₄ → 2 CO₂ + K₂O + 2 MnO₃ + H₂O
22.14g of KMnO₄ in 1,00L of water have a molarity of:
22,14g KMnO₄/L× = 0,14 M of KMnO₄
The moles you required for a complete reaction are:
0,14 mol KMnO₄/L× 0,02050L = 2,872x10⁻³ mol KMnO₄
By the reaction 2 moles of KMnO₄ reacts with 1 mole of H₂C₂O₄, thus, the moles of H₂C₂O₄ that react were:
2,872x10⁻³ mol KMnO₄× = 1,436x10⁻³ moles of H₂C₂O₄
As the volume of the sample was 200,0mL≡ 0,2000L, The concentration of H₂C₂O₄ is:
= <em>7,1800x10⁻³ M</em>
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I hope it helps!