The Henry's Law constant of methyl bromide, CH3Br, is k = 0.159 mol/(L ⋅ atm) at 25°C. What is the solubility of methyl bromide in water at 25°C and at a partial pressure of 265 mm Hg? The Henry's Law constant of methyl bromide, CH3Br, is k = 0.159 mol/(L atm) at 25°C. What is the solubility of methyl bromide in water at 25°C and at a partial pressure of 265 mm Hg? 0.0554 mol/L 0.355 mol/L 47.7 mol/L 0.483 mol/L

.055 mol/L

Explanation:

Henry's Law is about solubility of a gas in a liquid which is dependent on the partial pressure of gas above the liquid . According to it

C = k P where C is concentration of gas in the liquid , k is henry's constant and P is partial pressure of the gas .

Given k = .159 mol/L.atm

P = 265 mm of Hg

= 265 / 760

= .34868 atm  

C = kP

= .159 x .34868

= .055 mol/L

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.