SELECT * FROM question_mgmt as q WHERE id=1773 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 h2s a toxic gas with rotten egg like smell is us

Question:

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 = 1000g / 18g mol-1

= 55.56 mol

 

∴Mole fraction of H2S, x =  Moles of H2S / Moles of H2S + Moles of water

0.195 / (0.195 + 55.56)

= 0.0035

 

At STP, pressure (p) = 0.987 bar

According to Henry's law:

p= KHx

⇒ KH = p / x

= 0.0987 / 0.0035 bar

= 282 bar


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