How many atoms of iodine are in 12.75g of cai2

0.5  × 10²³ atoms of iodine

Explanation:

Given data:

Mass of calcium iodide = 12.75 g

Number of atoms of iodine = ?

Solution:

First of all we will calculate the number of moles of calcium iodide.

Number of moles = mass/ molar mass

Number of moles = 12.75 g/ 293.9 g/mol

Number of moles = 0.04 mol

In one mole of calcium iodide there are two moles of iodine.

Thus in 0.04 moles:

0.04 mol × 2 = 0.08 moles of iodine

Now we will use the Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol

0.5  × 10²³ atoms of iodine.

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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