# A chemist wishing to do an experiment requiring 47 Ca 2+ (half-life = 4.5 days) needs 5.0 μ g of the nuclide. What mass of 47 CaCO 3 must be ordered if it takes 48 h for delivery from the supplier? Assume that the atomic mass of 47 Ca is 47.0 u.

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 29E
Textbook Problem
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## A chemist wishing to do an experiment requiring 47Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47CaCO3 must be ordered if it takes 48 h for delivery from the supplier? Assume that the atomic mass of 47Ca is 47.0 u.

Interpretation Introduction

Interpretation: Mass of 47CaCO3 required for 5μg of the 47Ca2+ nuclide in the given experiment is to be stated.

Concept introduction: A process through which a unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. The cause of instability of a nuclide is its inefficiency in holding the nucleus together.

Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

The decay constant can be calculated by the formula given below.

λ=0.693t1/2

The time of decay can be calculated by the formula given below

t=2.303λlogn0n

To determine: Mass of 47CaCO3 required for 5μg of the 47Ca2+ nuclide in the given experiment.

### Explanation of Solution

Explanation

The integrated first order rate law is,

ln[NN0]=Ktln[NN0]=[0.693t1/2]×t

Where,

N is the mass of nuclide required.

N0 is the mass of nuclide to be ordered

t1/2 is the half life of nuclide.

t is the half life of the nuclide.

Substitute the value of N , t1/2 , t in the above equation.

ln[5.0μgN0]=[0.6934.5]×2ln[5.0μgN0]=0.31[5.0μgN0]=e0.31N0=6.8μg_

The amount of 47CaCO3 required is 15

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