Page No 41: - Chapter 2 - Solutions - Exercise Solutions - Ncert Solutions class 12 - Chemistry

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Question 2.6:

H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer:

It is given that the solubility of H₂S in water at STP is 0.195 m, i.e., 0.195 mol of H₂S is dissolved in 1000 g of water.

Moles of water 

= 55.56 mol

∴Mole fraction of H₂S, x

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law:

p = K_Hx

= 282 bar

Question 2.7:

Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Answer:

It is given that:

K_H = 1.67 × 10⁸ Pa

 = 2.5 atm = 2.5 × 1.01325 × 10⁵ Pa

= 2.533125 × 10⁵ Pa

According to Henry’s law:

= 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

= 27.78 mol of water

Now,

Hence, quantity of CO₂ in 500 mL of soda water = (0.042 × 44)g

= 1.848 g