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

explanation:

i think option no d is the correct answer

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%