Answer :
Sure, let's break this problem down step by step.
We have a solution that is 2 liters (L) in volume and is described as [tex]\( D 51 / 2 NS + 20 \text{ mEq KCl} \)[/tex]. We need to find out:
a. How many milligrams of dextrose are in the solution.
b. How many grams of normal saline (NS) are in the solution.
c. How many milliequivalents of potassium chloride (KCl) are in the solution.
### Part a: Milligrams of Dextrose
"D 5" means a 5% dextrose solution, by weight. This means that in every 100 grams of the solution, there are 5 grams of dextrose.
First, convert the percentage to a fraction:
[tex]\[ 5\% = \frac{5}{100} = 0.05 \][/tex]
Next, understand that 1 liter of water weighs approximately 1 kilogram (kg), which is 1000 grams (g), so a 5% solution has:
[tex]\[ 0.05 \times 1000 \text{ g} = 50 \text{ g of dextrose per liter} \][/tex]
Since we have 2 liters of this solution:
[tex]\[ 50 \text{ g/L} \times 2 \text{ L} = 100 \text{ g of dextrose total} \][/tex]
Now convert grams to milligrams (mg), knowing that 1 gram = 1000 milligrams:
[tex]\[ 100 \text{ g} \times 1000 \text{ mg/g} = 100,000 \text{ mg of dextrose} \][/tex]
Thus, the solution contains:
[tex]\[ \boxed{100,000 \text{ mg}} \][/tex]
### Part b: Grams of NS
"1/2 NS" means a half normal saline solution, with "normal saline" (NS) typically referring to a 0.9% sodium chloride (NaCl) solution. Therefore, 1/2 NS has a concentration of:
[tex]\[ 0.9\% \times \frac{1}{2} = 0.45\% \][/tex]
Convert the percentage to a fraction:
[tex]\[ 0.45\% = \frac{0.45}{100} = 0.0045 \][/tex]
The concentration of NS in grams per liter is:
[tex]\[ 0.0045 \times 1000 \text{ g} = 4.5 \text{ g/L} \][/tex]
Since we have 2 liters of the solution:
[tex]\[ 4.5 \text{ g/L} \times 2 \text{ L} = 9 \text{ g of NS total} \][/tex]
Thus, the solution contains:
[tex]\[ \boxed{9 \text{ g}} \][/tex]
### Part c: Milliequivalents of KCl
The problem directly states there are 20 milliequivalents (mEq) of potassium chloride (KCl) in the solution. Thus, we can directly write:
[tex]\[ \boxed{20 \text{ mEq}} \][/tex]
So, the detailed breakdown of the components in the 2-liter solution is as follows:
a. [tex]\(100,000 \text{ mg}\)[/tex] of dextrose
b. [tex]\(9 \text{ g}\)[/tex] of NS (sodium chloride)
c. [tex]\(20 \text{ mEq}\)[/tex] of KCl
We have a solution that is 2 liters (L) in volume and is described as [tex]\( D 51 / 2 NS + 20 \text{ mEq KCl} \)[/tex]. We need to find out:
a. How many milligrams of dextrose are in the solution.
b. How many grams of normal saline (NS) are in the solution.
c. How many milliequivalents of potassium chloride (KCl) are in the solution.
### Part a: Milligrams of Dextrose
"D 5" means a 5% dextrose solution, by weight. This means that in every 100 grams of the solution, there are 5 grams of dextrose.
First, convert the percentage to a fraction:
[tex]\[ 5\% = \frac{5}{100} = 0.05 \][/tex]
Next, understand that 1 liter of water weighs approximately 1 kilogram (kg), which is 1000 grams (g), so a 5% solution has:
[tex]\[ 0.05 \times 1000 \text{ g} = 50 \text{ g of dextrose per liter} \][/tex]
Since we have 2 liters of this solution:
[tex]\[ 50 \text{ g/L} \times 2 \text{ L} = 100 \text{ g of dextrose total} \][/tex]
Now convert grams to milligrams (mg), knowing that 1 gram = 1000 milligrams:
[tex]\[ 100 \text{ g} \times 1000 \text{ mg/g} = 100,000 \text{ mg of dextrose} \][/tex]
Thus, the solution contains:
[tex]\[ \boxed{100,000 \text{ mg}} \][/tex]
### Part b: Grams of NS
"1/2 NS" means a half normal saline solution, with "normal saline" (NS) typically referring to a 0.9% sodium chloride (NaCl) solution. Therefore, 1/2 NS has a concentration of:
[tex]\[ 0.9\% \times \frac{1}{2} = 0.45\% \][/tex]
Convert the percentage to a fraction:
[tex]\[ 0.45\% = \frac{0.45}{100} = 0.0045 \][/tex]
The concentration of NS in grams per liter is:
[tex]\[ 0.0045 \times 1000 \text{ g} = 4.5 \text{ g/L} \][/tex]
Since we have 2 liters of the solution:
[tex]\[ 4.5 \text{ g/L} \times 2 \text{ L} = 9 \text{ g of NS total} \][/tex]
Thus, the solution contains:
[tex]\[ \boxed{9 \text{ g}} \][/tex]
### Part c: Milliequivalents of KCl
The problem directly states there are 20 milliequivalents (mEq) of potassium chloride (KCl) in the solution. Thus, we can directly write:
[tex]\[ \boxed{20 \text{ mEq}} \][/tex]
So, the detailed breakdown of the components in the 2-liter solution is as follows:
a. [tex]\(100,000 \text{ mg}\)[/tex] of dextrose
b. [tex]\(9 \text{ g}\)[/tex] of NS (sodium chloride)
c. [tex]\(20 \text{ mEq}\)[/tex] of KCl
