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CO₂ is added to 1.00 L of pure water at 25°C in a pressurised bottle - VCE - SSCE Chemistry - Question 6 - 2004 - Paper 1

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CO₂ is added to 1.00 L of pure water at 25°C in a pressurised bottle. The pressure of CO₂ above the water was raised to 3.00 atm and the gaseous CO₂ came to equilibr... show full transcript

Worked Solution & Example Answer:CO₂ is added to 1.00 L of pure water at 25°C in a pressurised bottle - VCE - SSCE Chemistry - Question 6 - 2004 - Paper 1

Step 1

Mass of CO₂ dissolved in water

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Answer

First, we need to calculate the number of moles of CO₂ that is dissolved in water. Given that the mass of CO₂ is 5.00 g and its molar mass is approximately 44.01 g/mol, we can use the formula:

n(CO2)=massmolar mass=5.00 g44.01 g/mol0.1136 moln(CO_{2}) = \frac{mass}{molar\ mass} = \frac{5.00\ g}{44.01\ g/mol} \approx 0.1136\ mol

Step 2

Concentration of CO₂ in solution

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Answer

To find the concentration of CO₂ in the 1.00 L solution, we can use the formula:

[CO2(aq)]=n(CO2)Volume=0.1136 mol1.00 L=0.1136 M[CO_{2}(aq)] = \frac{n(CO_{2})}{Volume} = \frac{0.1136\ mol}{1.00\ L} = 0.1136\ M

Step 3

Pressure of CO₂

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The pressure of CO₂ is given as 3.00 atm. Therefore:

p(CO2,g)=3.00 atmp(CO_{2}, g) = 3.00\ atm

Step 4

Calculating the equilibrium constant (K_c)

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Answer

Now we can substitute these values into the equation for the equilibrium constant:

Kc=p(CO2,g)[CO2(aq)]=3.00 atm0.1136 M26.4 atm M1K_{c} = \frac{p(CO_{2}, g)}{[CO_{2}(aq)]} = \frac{3.00\ atm}{0.1136\ M} \approx 26.4\ atm\ M^{-1}

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