Chapter 1, Problem 1.58P

a)

Interpretation Introduction

Interpretation:Average mass for below sets of measurements is to be identified.

  $\begin{array}{l}\text{I: 8}\text{.72 g, 8}\text{.74 g, 8}\text{.70 g}\\ \text{II: 8}\text{.56 g, 8}\text{.77 g, 8}\text{.83 g}\\ \text{III: 8}\text{.50 g, 8}\text{.48 g, 8}\text{.51 g}\\ \text{IV: 8}\text{.41 g, 8}\text{.72 g, 8}\text{.55 g}\end{array}$

Concept introduction:Any test, trial or tentative procedure that is executed under highly controlled conditions whose purpose is to examine validity of any hypothesis is called an experiment. These are also performed so as to discover unknown things in nature.

The extent to which calculated or measured values are close to standard one is called accuracy.

Answer to Problem 1.58P

Average mass of set I is $8.72\text{ g}$ .

Average mass of set II is $8.72\text{ g}$ .

Average mass of set III is $8.50\text{ g}$ .

Average mass of set IV is $8.56\text{g}$ .

Set I and set II are the most accurate.

Explanation of Solution

The formula to calculate the average mass is as follows:

  $\text{Average mass}=\frac{M_1+M_2+M_3}{3}$

The value of $M_1$ is $8.72\text{ g}$ .

The value of $M_2$ is $8.74\text{ g}$ .

The value of $M_3$ is $8.70\text{ g}$ .

Substitute the values in above equation to calculate average mass for set I.

  $\begin{array}{c}\text{Average mass}=\frac{M_1+M_2+M_3}{3}\\ =\frac{8.72\text{ g}+8.74\text{ g}+8.70\text{ g}}{3}\\ =8.72\text{g}\end{array}$

The value of $M_1$ is $8.56\text{ g}$ .

The value of $M_2$ is $8.77\text{ g}$ .

The value of $M_3$ is $8.83\text{ g}$ .

Substitute the values in above equation to calculate average mass for set II.

  $\begin{array}{c}\text{Average mass}=\frac{M_1+M_2+M_3}{3}\\ =\frac{8.56\text{ g}+8.77\text{ g}+8.83\text{ g}}{3}\\ =8.72\text{g}\end{array}$

The value of $M_1$ is $8.50\text{ g}$ .

The value of $M_2$ is $8.48\text{ g}$ .

The value of $M_3$ is $8.51\text{ g}$ .

Substitute the values in above equation to calculate average mass for set III.

  $\begin{array}{c}\text{Average mass}=\frac{M_1+M_2+M_3}{3}\\ =\frac{8.50\text{g}+8.48\text{g}+8.51\text{g}}{3}\\ =8.50\text{g}\end{array}$

The value of $M_1$ is $8.41\text{ g}$ .

The value of $M_2$ is $8.72\text{ g}$ .

The value of $M_3$ is $8.55\text{ g}$ .

Substitute the values in above equation to calculate average mass for set IV.

  $\begin{array}{c}\text{Average mass}=\frac{M_1+M_2+M_3}{3}\\ =\frac{8.41\text{g}+8.72\text{g}+8.55\text{g}}{3}\\ =8.56\text{g}\end{array}$

Since set I and set II have values equal to that of standard value $\left(8.72\text{ g}\right)$ , both these sets are the most accurate.

b)

Interpretation Introduction

Interpretation:Among below sets of measurements, the most precise set should be determined. Also, whether this set is the most accurate or not should be determined.

  $\begin{array}{l}\text{I: 8}\text{.72 g, 8}\text{.74 g, 8}\text{.70 g}\\ \text{II: 8}\text{.56 g, 8}\text{.77 g, 8}\text{.83 g}\\ \text{III: 8}\text{.50 g, 8}\text{.48 g, 8}\text{.51 g}\\ \text{IV: 8}\text{.41 g, 8}\text{.72 g, 8}\text{.55 g}\end{array}$

Concept introduction:The extent to which calculated or measured values are close to standard one is called accuracy.The closeness of measurements with each other is termed as precision. It determines the extent to which measurements are close to each other.

Explanation of Solution

Precision is determined by range of data. The formula to calculate range is as follows:

  $\text{Range}=\left(\text{Largest value}\right)-\left(\text{smallest value}\right)$

The largest value is $8.74\text{ g}$ .

The smallest value is $8.70\text{ g}$ .

Substitute the values in above equation for range of set I.

  $\begin{array}{c}\text{Range}=\left(\text{Largest value}\right)-\left(\text{smallest value}\right)\\ =\left(8.74-8.70\right)\text{ g}\\ =0.04\text{ g}\end{array}$

The largest value is $8.83\text{ g}$ .

The smallest value is $8.56\text{ g}$ .

Substitute the values in above equation for range of set II.

  $\begin{array}{c}\text{Range}=\left(\text{Largest value}\right)-\left(\text{smallest value}\right)\\ =\left(8.83\text{g}\right)-\left(8.56\text{g}\right)\\ =0.27\text{g}\end{array}$

The largest value is $8.51\text{ g}$ .

The smallest value is $8.48\text{ g}$ .

Substitute the values in above equation for range of set III.

  $\begin{array}{c}\text{Range}=\left(\text{Largest value}\right)-\left(\text{smallest value}\right)\\ =\left(8.51\text{g}\right)-\left(8.48\text{g}\right)\\ =0.03\text{g}\end{array}$

The largest value is $8.72\text{ g}$ .

The smallest value is $8.41\text{ g}$ .

Substitute the values in above equation for range of set IV.

  $\begin{array}{c}\text{Range}=\left(\text{Largest value}\right)-\left(\text{smallest value}\right)\\ =\left(8.72\text{g}\right)-\left(8.41\text{g}\right)\\ =0.31\text{g}\end{array}$

Since range of set III comes out to be the least among all sets of measurements, it is the most precise measurement. But value of its average mass is quite different from that of standard value; it is not the most accurate measurement.

c)

Interpretation Introduction

Interpretation:Among below sets of measurements, set that is the most accurate and the most precise should be determined.

  $\begin{array}{l}\text{I: 8}\text{.72 g, 8}\text{.74 g, 8}\text{.70 g}\\ \text{II: 8}\text{.56 g, 8}\text{.77 g, 8}\text{.83 g}\\ \text{III: 8}\text{.50 g, 8}\text{.48 g, 8}\text{.51 g}\\ \text{IV: 8}\text{.41 g, 8}\text{.72 g, 8}\text{.55 g}\end{array}$

Concept introduction:The extent to which calculated or measured values are close to standard one is called accuracy. The closeness of measurements with each other is termed as precision. It determines the extent to which measurements are close to each other.

Explanation of Solution

Average value of set I match with that of standard value and its range is quite low so it is the most accurate and most precise set of measurement.

d)

Interpretation Introduction

Interpretation:Among below sets of measurements, set that is the least accurate and the least precise should be determined.

  $\begin{array}{l}\text{I: 8}\text{.72 g, 8}\text{.74 g, 8}\text{.70 g}\\ \text{II: 8}\text{.56 g, 8}\text{.77 g, 8}\text{.83 g}\\ \text{III: 8}\text{.50 g, 8}\text{.48 g, 8}\text{.51 g}\\ \text{IV: 8}\text{.41 g, 8}\text{.72 g, 8}\text{.55 g}\end{array}$

Concept introduction:The extent to which calculated or measured values are close to standard one is called accuracy. The closeness of measurements with each other is termed as precision. It determines the extent to which measurements are close to each other.

Explanation of Solution

Average value of set I differs highly from standard value and its range is maximum so it is the least accurate and least precise set of measurement.