Chapter 1, Problem 5MCP

(a)

Interpretation Introduction

Interpretation:

A volumetric flask contains $\text{50}\text{.0}\text{mL}$ of water.  One physical and chemical property of water in the flask has to be given.

Concept Introduction:

Physical properties:

The properties which are measured or observed without change in composition of matter are said to be physical properties.  These includes appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity and so on.

Chemical properties:

The properties that are observed when a substance undergoes any chemical reaction are said to be chemical properties.  These include flammability, toxicity, stability, heat of combustion, reactivity, acidity, basicity and so on.

Explanation of Solution

The physical property of water is said to be that water is colorless liquid.

The chemical property of water is the reactivity of water with substances.

(b)

Interpretation Introduction

Interpretation:

The properties that are given in part (a) has to be labelled as intensive or extensive properties.

Concept Introduction:

Extensive property:

The property which depends on the amount of the matter is said to be an extensive property.  The change in the volume or mass of the matter changes the extensive properties.

Intensive property:

The property which does not depends on the amount of matter is said to be an intensive property.  The change in the volume or mass of the matter does not affect intensive properties.

Explanation of Solution

The physical property of water that is colorless liquid is an intensive property.  It does not depend on the mass or volume of the water taken.

The reactivity of water with other substances is said to be extensive property.  It depends on the volume of water taken.

(c)

Interpretation Introduction

Interpretation:

A volumetric flask contains $\text{50}\text{.0}\text{mL}$ of water.  The mass of the water has to be calculated by the density of the water $\text{1}\text{.00}{\text{g/cm}}^{\text{3}}$.

Concept Introduction:

Density:

The ratio of mass of a sample to its volume is said to be the density.

  $\text{Density = }\frac{\text{mass}}{\text{volume}}$

Explanation of Solution

Given, volume of water is $\text{50}\text{.0}\text{mL}$ and density of water is $\text{1}\text{.00}{\text{g/cm}}^{\text{3}}$.

The mass of the water is calculated as

  $\begin{array}{l}\text{1}\text{mL}\text{=}\text{1}{\text{cm}}^{\text{3}}\\ \text{Mass}\text{=}\text{Density}\text{×}\text{volume}\\ \text{=}\text{1}\text{.00}{\text{g/cm}}^{\text{3}}\text{×}\text{50}\text{.0}{\text{cm}}^{\text{3}}\\ \text{=}\text{50}\text{.0}\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The mass of water calculated in part (c) has to be converted into ounces.

Concept Introduction:

The mass can be converted from grams into ounces by the conversion factor.

  $\begin{array}{l}\text{1}\text{lb}\text{=}\text{454}\text{g}\\ \text{Conversion}\text{factor:}\\ \text{=}\frac{\text{1}\text{lb}}{\text{454}\text{g}}\text{or}\text{=}\frac{\text{454}\text{g}}{\text{1}\text{lb}}\\ \text{1}\text{lb}\text{=}\text{16}\text{oz}\\ \text{=}\frac{\text{1}\text{lb}}{\text{16}\text{oz}}\text{or}\frac{\text{16}\text{oz}}{\text{1}\text{lb}}\end{array}$

Explanation of Solution

The mass of the water is calculated as $\text{50}\text{.0}\text{g}$.

It can be converted into ounces as follows:

  $\begin{array}{l}\text{=}\text{50}\text{.0}\text{g}\text{×}\frac{\text{1}\text{lb}}{\text{454}\text{g}}\text{×}\frac{\text{16}\text{oz}}{\text{1}\text{lb}}\\ \text{=}\text{1}\text{.76}\text{oz}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The mass of water calculated in part (c) has to be converted into pounds.

Concept Introduction:

The mass can be converted from grams into pounds by the conversion factor.

  $\begin{array}{l}\text{1}\text{lb}\text{=}\text{454}\text{g}\\ \text{Conversion}\text{factor:}\\ \text{=}\frac{\text{1}\text{lb}}{\text{454}\text{g}}\text{or}\text{=}\frac{\text{454}\text{g}}{\text{1}\text{lb}}\end{array}$

Explanation of Solution

The mass of the water is calculated as $\text{50}\text{.0}\text{g}$.

It can be converted into pounds as follows:

  $\begin{array}{l}\text{=}\text{50}\text{.0}\text{g}\text{×}\frac{\text{1}\text{lb}}{\text{454}\text{g}}\\ \text{=}\text{0}\text{.11}\text{lb}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Among the units that are calculated in part (d) and part (e), the appropriate units have to be given.

Explanation of Solution

The appropriate units for the mass is pounds.

The metric unit for mass is grams.  The grams can be directly converted into pounds which is the unit of English system.  To convert grams into ounce, it has to be converted to pounds.  So, pounds are the appropriate units.