Question

question 15 (2.5 points)
a 1 g vial of acyclovir that has been reconstituted to a concentration of 50 mgt/ml has enough for 3 doses of 0.3 g each.
true
false
question 16 (2.5 points)
multidose vials of claforan 2 g are diluted with 10 ml of normal saline to yield 180 mg/ml. how many ml of the medication should be given if the patient is prescribed to receive 2000 mg of the medication?
9.6 ml
10.0 ml
11.1 ml
12.4 ml
question 17 (2.5 points)
given that the percentage strength of normal saline is 0.9%, how many grams of sodium chloride is in 1 l of normal saline?
9 g
0.9 g
90 mg
900 mg

Explanation:

Question 15

Step 1: Convert units

First, convert the concentration to g/mL. Since \( 1\space g = 1000\space mg \), \( 50\space mg/mL=\frac{50}{1000}\space g/mL = 0.05\space g/mL \).

Step 2: Find total volume

The vial has \( 1\space g \) of acyclovir. Using the formula \( Volume=\frac{Mass}{Concentration} \), the volume of the reconstituted solution is \( \frac{1\space g}{0.05\space g/mL}=20\space mL \).

Step 3: Calculate total dose volume

Each dose is \( 0.3\space g \). The volume for one dose is \( \frac{0.3\space g}{0.05\space g/mL} = 6\space mL \). For 3 doses, the total volume is \( 3\times6\space mL = 18\space mL \), which is less than \( 20\space mL \). So the statement is True.

Step 1: Recall the formula

We use the formula \( Volume=\frac{Mass}{Concentration} \). Here, the mass needed is \( 2000\space mg \) and the concentration is \( 180\space mg/mL \).

Step 2: Calculate the volume

Substitute the values into the formula: \( Volume=\frac{2000\space mg}{180\space mg/mL}\approx11.1\space mL \).

Step 1: Understand percentage strength

A \( 0.9\% \) strength means \( 0.9\space g \) of sodium chloride per \( 100\space mL \) of solution (since percentage strength \( \%w/v=\frac{Mass\space of\space solute\space (g)}{Volume\space of\space solution\space (mL)}\times100 \)).

Step 2: Calculate for 1 L

Since \( 1\space L = 1000\space mL \), we set up a proportion. Let \( x \) be the mass of sodium chloride in \( 1000\space mL \). Then \( \frac{0.9\space g}{100\space mL}=\frac{x}{1000\space mL} \). Solving for \( x \), we get \( x=\frac{0.9\space g\times1000\space mL}{100\space mL}=9\space g \).

Answer:

True

Question 16