Chapter 1, Problem 1.96P

Interpretation Introduction

(a)

Interpretation:

The amount of fluid received by the patient in $20\text{ min}$ is to be calculated.

Concept introduction:

The conversion factor to convert minutes to hours is as follows:

  $1\text{ hour}=60\text{ min}$

Answer to Problem 1.96P

The volume of fluid received by the patient in $20\text{ min}$ is $50\text{ mL}$.

Explanation of Solution

The given rate of flow of intravenous solution is represented as follows:

  $\frac{150\text{ mL}}{1\text{ hour}}$

The conversion factor to convert minutes to hours is as follows:

  $1\text{ hour}=60\text{ min}$

Hence the factor to convert $20\text{ min}$ to hours as follows:

  $\begin{array}{r}\text{Time taken}=\left(20\min \right)\left(\frac{1\text{ hour}}{60\min }\right)\\ =\frac{1}{3}\text{ hours}\end{array}$

So, the fluid received by the patient is calculated as follows:

  $\begin{array}{c}\text{Volume of fluid}=\left(\frac{150\text{ mL}}{1\text{ hour}}\right)\left(\frac{1}{3}\text{hours}\right)\\ =50\text{ mL}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The duration in which the patient would receive $90\text{ mL}$ of fluid is to be calculated.

Concept introduction:

The conversion factor to convert the milligrams to grams is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Answer to Problem 1.96P

The duration in which the patient would receive $90\text{ mL}$ of fluid is $36\text{ min}$.

Explanation of Solution

Since the amount of fluid received by the patient in $20\text{ min}$ is $50\text{ mL}$ . It can be represented as follows:

  $\frac{20\text{ min}}{50\text{ mL}}$

So, the duration in which the patient would receive $90\text{ mL}$ of fluid is calculated as follows:

  $\begin{array}{c}\text{Time taken by patient}=\left(\frac{20\text{ min}}{50\text{ mL}}\right)\left(90\text{ mL}\right)\\ =36\text{ min}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The duration in which the bag that has a volume of $600\text{ mL}$ is emptied is to be determined.

Concept introduction:

The conversion factor to convert the milligrams to grams is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Answer to Problem 1.96P

The duration in which the bag is emptied is $4\text{ hours}$.

Explanation of Solution

The rate of flow of intravenous solution can be represented as follows:

  $\frac{1\text{ hour}}{150\text{ mL}}$

So, the duration in which the bag is emptied is calculated as follows:

  $\begin{array}{c}\text{Time in which bag emptied}=\left(\frac{1\text{ hour}}{150\text{ mL}}\right)\left(600\text{ mL}\right)\\ =4\text{ hours}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The duration in which $2.0\text{ g}$ glucose is administered to the patient is to be determined.

Concept introduction:

The conversion factor to convert the milligrams to grams is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Answer to Problem 1.96P

The duration in which $2.0\text{ g}$ glucose is administered to the patient is $9\min$.

Explanation of Solution

The conversion factor to convert the milligrams to grams is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Hence the factor to convert $\text{90 mg}$ of glucose to $\text{g}$ as follows:

  $\begin{array}{c}\text{Mass of glucose}=\left(\text{90 mg}\right)\left(\frac{{10}^{-3}\text{g}}{\text{1}\text{mg}}\right)\\ =90\times {10}^{-3}\text{ g}\end{array}$

Since $1\text{ mL}$ has $90\times {10}^{-3}\text{ g}$ , this can be represented as follows:

  $\frac{1\text{ mL}}{90\times {10}^{-3}\text{ g}}$

The volume equivalent to $2.0\text{ g}$ glucose is calculated as follows:

  $\begin{array}{c}\text{Volume of glucose}=\left(\frac{1\text{ mL}}{90\times {10}^{-3}\text{ g}}\right)\left(2.0\text{ g}\right)\\ =22.22\text{ mL}\end{array}$

Since the amount of fluid received by the patient in $20\text{ min}$ is $50\text{ mL}$ so, the duration in which $22.22\text{ mL}$ glucose is administered to the patient is calculated as follows:

  $\begin{array}{c}\text{Time taken by patient}=\left(\frac{20\min }{50\text{ mL}}\right)\left(22.22\text{mL}\right)\\ =8.888\text{ min}\\ \approx 9\min \end{array}$