Chapter 2, Problem 101AP

Make the following conversions.

1. 1 .25 in. to feet and to centimeters
2. 2.12 qt to gallons and to liters
3. 2640 ft to miles and to kilometers
4. 1.254 kg lead to its volume in cubic centimeters
5. 250. mL ethanol to its mass in grams
6. 3.5 in.3 of mercury to its volume in milliliters and its mass in kilograms

Interpretation Introduction

(a)

Interpretation:

Convert 1.25 in. to feet and to centimetres.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

1.25 in. is equal to $0.10\text{41667 ft}$

1.25 in. is equal to $\text{3}\text{.175 cm}$.

Explanation of Solution

To solve the problem, following conversion factors are used:

$1\text{ in}\text{.}=\text{0}\text{.0833333 ft}$

Here, ft is foot and in is inch.

$1\text{ in}=\text{2}\text{.54 cm}$

Here, in is inch and cm is centimetre

Convert 1.25 in. to feet:

Since, $1\text{ in}\text{.}=\text{0}\text{.0833333 ft}$

Therefore, 1.25 in. in feet

= $\frac{\text{0}\text{.0833333 ft}}{\text{in}\text{.}}\text{×1}\text{.25 in}\text{.}$ = $0.1041667\text{f}\text{t}$

Convert 1.25 in. to cm:

Since, $1\text{ in}=\text{2}\text{.54 cm}$

Therefore, 1.25 in. in cm

= $\frac{\text{2}\text{.54 cm}}{\text{in}\text{.}}\text{×1}\text{.25 in}\text{.}$ = $\text{3}\text{.175 cm}$.

Interpretation Introduction

(b)

Interpretation:

Convert 2.12 qt to gallons and to liters.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

2.12 qt is equal to $\text{0}\text{.53 gallon}$

2.12 qt is equal to $\text{2}\text{.00626836 liter}$.

Explanation of Solution

To solve the problem, following conversion factors are used:

$1\text{ qt}=\text{0}\text{.25 gallon}$

$1\text{ qt}=\text{0}\text{.946353 liter}$

$$

Here, qt is quart

Convert 2.12 qt to gallon:

Since, $1\text{ qt}=\text{0}\text{.25 gallon}$

Therefore, 2.12 qt in gallon

= $\frac{\text{0}\text{.25 gallon}}{\text{qt}}\text{×2}\text{.12 qt}$ = $\text{0}\text{.53 gallon}$

Convert 2.12 qt to liter:

Since, $\text{1 qt=0}\text{.946353 liter}$

Therefore, 2.12 qt in liter

= $\frac{\text{0}\text{.946353 liter}}{\text{qt}}\text{×2}\text{.12 qt}$ = $\text{2}\text{.00626836 liter}$.

Interpretation Introduction

(c)

Interpretation:

Convert 2640 ft to miles and to kilometres.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

2640 ft is equal to $0.5\text{m}\text{i}\text{l}\text{e}$

2640 ft is equal to $\text{0}\text{.804672 km}$.

Explanation of Solution

To solve the problem, following conversion factors are used:

$1\text{ ft}=\text{0}\text{.000189394 mile}$

$1\text{ ft}=\text{0}\text{.0003048 km}$

$$

Here, ft is feet and km is kilometre

Convert 2640 ft to miles:

Since, $1\text{ ft}=\text{0}\text{.000189394 mile}$

Therefore, 2640 ft in miles

= $\frac{\text{0}\text{.000189394 mile}}{\text{ft}}\text{×2640 ft}$ = $\text{0}\text{.5 mile}$

Convert 2640 ft to kilometres:

Since, $1\text{ ft}=\text{0}\text{.0003048 km}$

Therefore, 2640 in km

= $\frac{\text{0}\text{.0003048 km}}{\text{ft}}\text{×2640 ft}$ = $\text{0}\text{.804672 km}$.

Interpretation Introduction

(d)

Interpretation:

Volume of lead in cubic centimetres having 1.254 kg should be calculated.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

Volume of lead = 110.58 cm³.

Explanation of Solution

Since, density of lead is equal to 11.34 g/cm³

Density is equal to the ratio of mass to the volume.

Density = $\frac{\text{Mass}}{\text{Volume}}$

Convert mass into g:

Since, 1 kg = 1000 g

Mass of lead = ( $\text{1}\text{.254 }\times \text{1000}$ ) g = 1254 g

Put the value of mass and density in above expression, we get

11.34 g/cm³ = $\frac{1254\text{ g}}{\text{Volume}}$

Volume = $\frac{1254\text{ g}}{11.34{\text{ g/cm}}^3}$ = 110.58 cm³.

Interpretation Introduction

(e)

Interpretation:

Mass of ethanol in grams having 250. mL volume should be calculated.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

Mass of ethanol = 196.25 g.

Explanation of Solution

Since, density of ethanol is equal to 0.785 g/cm³

Density is equal to the ratio of mass to the volume.

Density = $\frac{\text{Mass}}{\text{Volume}}$

Convert volume into cm³ :

Since, 1 mL = 1 cm³

Volume of ethanol in cm³ = 250. cm³

Put the value of volume and density in above expression, we get

0.785 g/cm³ = $\frac{\text{ mass}}{250.{\text{ cm}}^{\text{3}}}$

Mass = $250.{\text{ cm}}^3\times 0.785{\text{ g/cm}}^3$ = 196.25 g.

Interpretation Introduction

(f)

Interpretation:

Volume of 3.5 in.3 of mercury in milliliters and its mass in kilograms should be calculated.

Concept Introduction:

A conversion factor is a numerical ratio which is used to show a measurement in one unit as another unit. It is always equal to 1.

Answer to Problem 101AP

Volume of mercury in mL is 57.3547 mL and its mass in kg is 0.78002 kg.

Explanation of Solution

Since, 1 in.3 = 16.3871 mL

Therefore, 3.5 in.3 = 57.3547 mL

Volume in mL = 57.3547 mL

Now, density of mercury = 13.6 g/cm³

Convert volume into cm³

Since, 1 cm³ = 1 mL

So, volume in cm³ = 57.3547 cm³

Density = $\frac{\text{Mass}}{\text{Volume}}$

Put the value of volume and density in above expression, we get

13.6 g/cm³ = $\frac{\text{ mass}}{57.3547{\text{ cm}}^{\text{3}}}$

Mass = $57.3547{\text{ cm}}^3\times 13.6{\text{ g/cm}}^3$ = 780.02 g

As, 1 kg = 1000 g

Thus, mass in kg = $\frac{780.02}{1000}$ = 0.78002 kg.