How many ml of 50% dextrose should be added to 1 L of 0.9% NaCl to make a 5% dextrose solution?
4 answers
I am wondering if your percents are by mass, or what?
I don't know it never said in the problem.
It is understood to be mass/volume for ease of dosage calculations.
Therefore whether it is 0.9% saline or simply water, the proportions are the same.
You can do the calculations as follows:
Suppose we need v ml of c1% concentrated dextrose diluted with 1000 ml of liquid to give a c2% solution,
mass of glucose added = v(c1/100) g
total volume of liquid = 1000+a
Therefore
(c2/100) = (v c1/100)/(1000+v)
Solving for v
(1000+v)c2 = v c1
v=1000c2/(c1-c2) ....(1)
For the case in point,
x=50, y=5
v=1000*5/(50-5)
= 111.1 ml.
Equation 1 is a handy formula to remember if you need to do the same calculation often and fast.
Therefore whether it is 0.9% saline or simply water, the proportions are the same.
You can do the calculations as follows:
Suppose we need v ml of c1% concentrated dextrose diluted with 1000 ml of liquid to give a c2% solution,
mass of glucose added = v(c1/100) g
total volume of liquid = 1000+a
Therefore
(c2/100) = (v c1/100)/(1000+v)
Solving for v
(1000+v)c2 = v c1
v=1000c2/(c1-c2) ....(1)
For the case in point,
x=50, y=5
v=1000*5/(50-5)
= 111.1 ml.
Equation 1 is a handy formula to remember if you need to do the same calculation often and fast.
I don't think adding salt would matter. The only one thing you're focusing on is how to make that 5% dextrose solution.
(x mL)*(50) + 1000 mL = (1000 mL + x) mL)*(5)
(x mL)*(50) + 1000 mL = (1000 mL + x) mL)*(5)
4 answers