Question

Saline solutions (NaCl in water) used to deliver intravenous drugs are 0.89%(w/v). What mass of NaCl would be needed to prepare 450 mL of such a solution?

A) 4 x 102 g
B) 0.89 g
C) 4.0 g
D) 5.1 g

Answer 1

To prepare 450 mL of a 0.89% saline solution, you would need 4.0 g of NaCl. This is calculated by using the percentage concentration to set up a proportion and then solving for the mass of NaCl.

Step-by-step explanation:

The question asks to calculate the mass of sodium chloride (NaCl) needed to prepare a 0.89% saline solution, which is commonly used for intravenous injections. The percentage indicates that there are 0.89 grams of NaCl per 100 mL of solution. To find the mass of NaCl needed for 450 mL of this solution, you would set up a proportion based on the percentage concentration:

0.89 g NaCl / 100 mL solution = x g NaCl / 450 mL solution

By cross-multiplying and solving for x, you get:

x = (0.89 g/100 mL) × 450 mL

x = 4.005 g NaCl

Therefore, you would need approximately 4.0 g of NaCl to prepare 450 mL of a 0.89% saline solution.

The correct answer to the question is Option C, 4.0 g.