3 persons solve it with 3 different answers!!!!!
available stock is 5 liter bag with conc.110 m.mol/L Na.Cl, and the Patient requires 140 m.mol/L for the 5 liter bag
How many ml of Na.Cl 0.3% should be added to the solution for the required Conc.?
(0.9 % Na.Cl = 514 m.mol/L)
Stock : 5 liter, 110 m moll L, therefore total of 550 m mol in 5 liter.
We need : 5 liter, 140 m mol/L, therefore total of 700 m mol in 5 liter.
We have : 0.3 % Nacl. = 171.33 m mol / L . (from conversion equation give for 0.9% and m mol relation for Nacl)
So Ultimately,
We have to find : Thus, we need 150 m mol extra, for to make a stock for our use.
Thus, 171.33 m mol----------------------------->1000 ml
Then, 150 m mol-----------------------------------> ?? ml
So 150 X 1000 / 171.33
= 875.50 ml of the given solution required to make stock for our use.
Ao answer should be ( must be!! ) 875,50 ml.
Is this right?!
Stock : 5 liter, 110 m moll L, therefore total of 550 m mol in 5 liter.
We need : 5 liter, 140 m mol/L, therefore total of 700 m mol in 5 liter.
We have : 0.3 % Nacl. = 171.33 m mol / L . (from conversion equation give for 0.9% and m mol relation for Nacl)
So Ultimately,
We have to find : Thus, we need 150 m mol extra, for to make a stock for our use.
Thus, 171.33 m mol----------------------------->1000 ml
Then, 150 m mol-----------------------------------> ?? ml
So 150 X 1000 / 171.33
= 875.50 ml of the given solution required to make stock for our use.
Ao answer should be ( must be!! ) 875,50 ml.
Is this right?!
130 ml