Question

complete parts (a) and (b).
a. a 5% dextrose solution (d5w) contains 5 mg of dextrose per 100 ml of solution. how many milligrams of dextrose is in 2500 ml of a 5% solution?
125 mg (round to the nearest tenth as needed.)
how many milliliters of d5w solution should be given to a patient needing 60 mg of dextrose?
1200 ml (round to the nearest milliliter as needed.)
b. normal saline solution (ns) has a concentration of 0.9% sodium chloride, or 0.9 mg per 100 ml. how many milligrams of sodium chloride is in 1.1 l of ns?
□ mg (round to the nearest tenth as needed.)

Explanation:

Step1: Convert volume to mL

Since 1 L = 1000 mL, then 1.1 L = 1.1×1000 = 1100 mL.

Step2: Set up proportion

Let $x$ be the amount of sodium - chloride in 1100 mL. We know that the concentration is 0.9 mg per 100 mL, so the proportion is $\frac{0.9}{100}=\frac{x}{1100}$.

Step3: Cross - multiply and solve for $x$

Cross - multiplying gives us $100x=0.9\times1100$. Then $100x = 990$, and $x=\frac{990}{100}=9.9$ mg.

Answer:

9.9