Answer :
Final answer:
The question pertains to calculating the volume of 0.9% sterile sodium chloride solution required based on body weight in pounds. After converting body weight to kilograms, finding the total meq required, and relating this to the mass of NaCl needed, we calculate the equivalent volume of a 0.9% NaCl solution. The exact calculation does not match with the provided options, indicating a possible discrepancy in the provided details.
Explanation:
The question involves calculating the volume of a 0.9% sterile sodium chloride solution needed for a person based on their body weight in meq/kg. First, we need to convert the weight from pounds to kilograms, knowing that 1 lb is approximately 0.453592 kg. After we have the person's weight in kilograms, we calculate the milliequivalents (meq) of sodium chloride required. Then, using the information that 1 liter of a 0.9% NaCl solution contains 9 grams of NaCl, and knowing that the molecular weight of NaCl is 58.44 g/mol, we can find the necessary volume of the solution in milliliters.
First, convert the weight:
187 lb * 0.453592 = 84.8 kg approximately.
Calculate total meq of NaCl needed:
84.8 kg * 3 meq/kg = 254.4 meq
Next, calculate grams of NaCl:
Because 1 meq of NaCl is equivalent to 1 mmol of NaCl, and 1 mmol of NaCl = 0.05844 g, we find the total grams by multiplying the meq by the weight of 1 mmol of NaCl.
254.4 meq * 0.05844 g/meq = 14.87 g of NaCl required.
Now, find the volume of 0.9% solution containing 14.87 g of NaCl:
A 0.9% solution means there are 9 g of NaCl per 1000 mL (1 liter). Therefore, we use a proportion to find the volume:
9 g/1000 mL = 14.87 g/X mL
X = (14.87 g * 1000 mL)/9 g
X = 1652.2 mL approximately.
However, this calculation does not match any of the provided options (a-d) because it seems there is an issue with the original question as it likely omitted the description of the concentration of the provided 0.9% sterile sodium chloride solution in mEq/L.
