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zepelin [54]
3 years ago
11

In the reaction: C6H6 + O2 = CO2 + H2O If 26.2 grams of C6H6 is reacted in a 0.250-liter container and this reaction is carried

out at 395 K, what is the pressure inside the container?
Chemistry
1 answer:
Lelechka [254]3 years ago
6 0
<h3>Answer:</h3>

43.33 atm

<h3>Explanation:</h3>

We are given;

Mass of C₆H₆ = 26.2 g

Volume of the container = 0.25 L

Temperature = 395 K

We are required to calculate the pressure inside the container;

First, we calculate the number of moles of C₆H₆

Molar mass of C₆H₆ =  78.1118 g/mol.

But; Moles = mass ÷ Molar mass

Moles of  C₆H₆ = 26.2 g ÷  78.1118 g/mol.

                         = 0.335 moles C₆H₆

Second, we calculate the pressure, using the ideal gas equation;

Using the ideal gas equation, PV = nRT , Where R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore;

P = nRT ÷ V

  = (0.335 mol × 0.082057 × 395 K) ÷ 0.25 L

 = 43.433 atm

Therefore, the pressure inside the container is 43.33 atm

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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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