Question

Problem 2.2 : The Henry's law constant of methyl bromide (CH Br), is 0.159 mol L^-1 bar^-1 at 25°C. What is the solubility of methyl bromide in water at 25°C and at pressure of 130 mmHg?
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Answer 1

Answer:

Explanation:

       

Given :-

Henry's law constant of CH Br

K = 0.159 $mol L^{-1}$$bar^{-1}$

Temperature = 25°C

Partial pressure (P) =  130 mm Hg               {Let us convert this to atm ot bar}

P = $\frac{130}{760}$ = 0.171 bar                  {to get approx value of 130 in terms of bar

                                                                  divide 130 by 760}

                                                                     

To Find:-

Solubility (S) of CH Br = ?

Solution:-

Solubility = K * P

               = 0.159 $mol L^{-1}$$bar^{-1}$ *  0.171 bar  

Solubility = 0.02719 $mol L^{-1}$

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Answer 2

Answer:

0.271 M

Explanation:

Henry's law Constant (KH)= 0.159 mol/L

P= 0.173 bar

Formula:- S= KH.P

S= 0.159×0.173

S=0.271M