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kaheart [24]
3 years ago
8

A 10 liter flask at 298 K contains a gaseous mixture of O2 and CO2 at 1 atmosphere. Which statement is true for the partial pres

sures of O2 and CO2 if 0.2 mole of O2 is present in the flask? (Given the universal gas constant R = 0.082 L∙atm/K∙mol)
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

The partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

Explanation:

From the Questions we are given;

Volume = 10 Liter

Temperature = 298 K

Pressure = 1 atm

We need to calculate the partial pressures of O₂ and CO₂

Step 1 : Number of moles of gaseous mixture

Using the ideal gas equation;

PV =nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature in K and R is the ideal gas constant (0.082 L∙atm/K∙mol)

Therefore;

n =\frac{PV}{RT}

n = \frac{(10)(1)}{(298)(0.082)}

Solving for n

n = 0.409 moles

Step 2: Moles of CO₂

Total number of moles of the mixture = 0.409 moles

Moles of Oxygen = 0.2 moles

Therefore;

Moles of CO₂ = 0.409 moles - 0.2 moles

                      = 0.209 moles

Step 3: Partial pressures of O₂ and CO₂

Partial pressure = \frac{No. of moles}{Total number of moles}(total pressure)

Therefore;

Partial pressure of Oxygen gas

= \frac{number of moles of Oxygen}{Total number of moles} (Total pressure)

= \frac{0.2}{0.409}(1)\\= 0.489 atm

Partial pressure of CO₂

= \frac{number of moles of CO₂}{Total number of moles} (Total pressure)

= \frac{0.209}{0.409}(1)\\= 0.511 atm

Thus, the partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

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The answer for the following problem has been mentioned below.

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Explanation:

Given:

volume of oxygen (V) = 4.50 L

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pressure of oxygen (P) = 2.50 atm

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To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

     P × V = n × R × T

As we know;

no of moles = \frac{m}{M}

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M represents the Gram molecular mass (M)

According to above mentioned equation;

           P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

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Substituting the values in the above equation;

                  2.50 × 4.50 = \frac{m}{16.0} × 0.0821 × 425

                   11.25 =  \frac{m}{16.0} × 34.8925

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Now;

As we know;

           \frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }

where;

m_{1} represents the mass of the oxygen

M_{1} represents the gram molecular mass of the oxygen

m_{2} represents the mass of the water

M_{2} represents the gram molecular mass of water

    From the above given formula,

      \frac{5.158}{16.0} = \frac{m_{2} }{18}

where;

Gram molecular weight of water = 18.0 u

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<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

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