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

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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