SELECT * FROM question_mgmt as q WHERE id=1774 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Solutions henry 39 s law constant for co2 in water is 1 67

Question:

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.

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


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