Photo AI

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

Question icon

Question 6

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.png

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

Calculate the number of moles of CO₂

96%

114 rated

Answer

To find the number of moles of CO₂ dissolved, we use the formula:

n(CO2)=massmolarmassn(CO_2) = \frac{mass}{molar \\ mass}

The molar mass of CO₂ is 44.01 g/mol:

n(CO2)=5.00g44.01g/mol0.1136moln(CO_2) = \frac{5.00 \, g}{44.01 \, g/mol} \approx 0.1136 \, mol

Step 2

Calculate the concentration of CO₂ in the solution

99%

104 rated

Answer

The concentration of CO₂ in the 1.00 L of solution is given by:

[CO2(aq)]=n(CO2)volume=0.1136mol1.00L=0.1136M[CO_2(aq)] = \frac{n(CO_2)}{volume} = \frac{0.1136 \, mol}{1.00 \, L} = 0.1136 \, M

Step 3

Calculate the pressure of CO₂ using the ideal gas law

96%

101 rated

Answer

Using the ideal gas law, we can find the pressure:

p(CO2)=nRTVp(CO_2) = \frac{nRT}{V}

Where R=0.0821LatmK1mol1R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}, T=273+25=298KT = 273 + 25 = 298 \, K, and V=1.00LV = 1.00 \, L:

p(CO2)=0.1136mol×0.0821L atmK1mol1×298 K1.00L2.78 atmp(CO_2) = \frac{0.1136 \, mol \times 0.0821 \, L \ atm \, K^{-1} \, mol^{-1} \times 298 \ K}{1.00 \, L} \approx 2.78 \ atm

Step 4

Calculate the equilibrium constant K_eq

98%

120 rated

Answer

Now, we can use the values obtained to calculate the equilibrium constant:

Keq=p(CO2, g)[CO2(aq)]=3.00atm0.1136M26.4atm M1K_{eq} = \frac{p(CO_2, \ g)}{[CO_2(aq)]} = \frac{3.00 \, atm}{0.1136 \, M} \approx 26.4 \, atm \ M^{-1}

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;