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? 9.9 mg (round to the nearest tenth as needed.)
how many milliliters of ns should be given to a patient needing 245 mg of sodium chloride?
ml (round to the nearest milliliter as needed.)

Explanation:

Step1: Set up proportion for first - part of b

The concentration of NS is 0.9 mg per 100 mL. Let $x$ be the volume of NS in mL for 245 mg of sodium chloride. The proportion is $\frac{0.9}{100}=\frac{245}{x}$.

Step2: Cross - multiply

Cross - multiplying gives $0.9x = 245\times100$. So, $0.9x=24500$.

Step3: Solve for $x$

Dividing both sides by 0.9, we get $x=\frac{24500}{0.9}\approx27222.22$. Rounding to the nearest milliliter, $x = 27222$ mL.

Answer:

27222