You want to make 100 ml of a 2.5 m stock solution of calcium chloride (molecular weight 110.98 g/mol). how many grams of calcium chloride do you need to weigh out in order to make this solution? round your answer to the nearest tenth of a gram.

27.7 g

Explanation:

To solve this problem we use the definition of molarity:

From the problem we're given M = 2.5 M and volume = 0.100 L. We use the above formula and calculate the required moles of calcium chloride (CaCl₂):

Finally we use the molecular weight of calcium chloride to calculate the required mass:

Rounding to the nearest tenth of a gram the answer is 27.7 g.

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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.