Henry's law constant for CO2 in water is 1.67 x 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
It is given that:
KH = 1.67 × 108Pa
PCO2 = 2.5 atm = 2.5 × 1.01325 × 105Pa
= 2.533125 × 105Pa
According to Henry's law:
PCO2 = KHX
⇒ x = PCO2 / KH
= 2.533125 × 105 / 1.67 × 108
= 0.00152
We can write, 
[Since, is negligible as compared to]
In 500 mL of soda water, the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write:
500 mL of water = 500 g of water
=500 / 18 mole of water
= 27.78 mol of water
Now, nCO2 / nH2O = x
nCO2 / 27.78 = 0.00152
nCO2 = 0.042 mol
Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44) g
= 1.848 g
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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It is given that:
KH = 1.67 × 108Pa
PCO2 = 2.5 atm = 2.5 × 1.01325 × 105Pa
= 2.533125 × 105Pa
According to Henry's law:
PCO2 = KHX
⇒ x = PCO2 / KH
...
Step-by-step explanation:
• It is given that:
•
• KH = 1
• 67 × 108Pa
•
Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.
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