Henry's law constant for CO, in water is 1.67x108 Pa at 298 K. Calcula

1.	Henry's law constant for CO, in water is 1.67x108 Pa at 298 K. Calculatethe quantity of Co, in 500 mL of soda water when packed under 2.5 atmCO, pressure at 298 K.
Answer» Answer: 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 Explanation: HOPE this helps you if it was helpful then please mark me as the brainliest and please follow

Henry's law constant for CO, in water is 1.67x108 Pa at 298 K. Calculatethe quantity of Co, in 500 mL of soda water when packed under 2.5 atmCO, pressure at 298 K.

Answer»

Answer:

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

Explanation:

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