Chapter 13, Problem 108A

Interpretation Introduction

(a)

Interpretation:

The unit gram needs to be converted into kilogram.

Concept introduction:

The concept of unit conversion is to be used.

Answer to Problem 108A

$\text{247}\text{g=0}\text{.247}\text{kg}$

Explanation of Solution

We know that

$\begin{array}{l}\text{1}\text{kg=1000g}\\ \text{or}\\ \text{1=}\frac{\text{1}\text{kg}}{\text{1000g}}\\ \therefore \text{247g=247g×}\frac{\text{1}\text{kg}}{\text{1000g}}\\ \text{=0}\text{.247}\text{kg}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The unit milligram needs to be converted into kilogram.

Concept introduction:

The concept of unit conversion is to be used.

Answer to Problem 108A

$\text{53}\text{mg=5}{\text{.3×10}}^{\text{-5}}\text{kg}$

Explanation of Solution

Since,

$\begin{array}{l}1kg=1000g\\ or\\ 1=\frac{1kg}{1000g}\\ and\\ 1g=1000mg\\ or\\ 1=\frac{1g}{1000mg}\\ \therefore 53mg=53mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}\\ =5.3\times 10^{-5}kg\end{array}$

Interpretation Introduction

(c)

Interpretation:

The unit milligram needs to be converted into kilogram.

Concept introduction:

The concept of unit conversion is to be used.

Answer to Problem 108A

$7.23\text{mg=7}{\text{.23×10}}^{\text{-6}}\text{kg}$

Explanation of Solution

We know that

$\begin{array}{l}1kg=1000g\\ or\\ 1=\frac{1kg}{1000g}\\ and\\ 1g=1000mg\\ or\\ 1=\frac{1g}{1000mg}\\ \therefore 7.23mg=7.23mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}\\ =7.23\times 10^{-6}kg\end{array}$

Interpretation Introduction

(d)

Interpretation:

The unit milligram needs to be converted into kilogram.

Concept introduction:

The concept of unit conversion is to be used.

Answer to Problem 108A

$975\text{mg=9}{\text{.75×10}}^{\text{-4}}\text{kg}$

Explanation of Solution

We know that

$\begin{array}{l}1kg=1000g\\ or\\ 1=\frac{1kg}{1000g}\\ and\\ 1g=1000mg\\ or\\ 1=\frac{1g}{1000mg}\\ \therefore 975mg=975mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}\\ =9.75\times 10^{-4}kg\end{array}$