How many grams of NaCl should be added to 1 L of water to make an isotonic solution? (molecular weight of NaCl= 58.5)
Can someone please explain to me why the answer is 9 grams.
To make an isotonic solution, you need to add enough NaCl to create a concentration equivalent to the concentration of solutes in the cell or tissue you want to mimic. In biological systems, the typical concentration of solutes is around 0.9%.
To calculate the number of grams of NaCl needed to make an isotonic solution in 1 L of water, you can use the following steps:
Step 1: Convert the desired concentration from a percentage to a decimal.
0.9% = 0.009 (0.9 divided by 100)
Step 2: Calculate the number of moles of NaCl needed.
To calculate the number of moles, divide the desired concentration by the molecular weight of NaCl.
0.009 / 58.5 = 0.000153 moles
Step 3: Convert moles to grams.
Multiply the number of moles by the molecular weight of NaCl to get the number of grams.
0.000153 moles x 58.5 g/mol = 0.00895 grams
So, approximately 0.009 grams of NaCl should be added to 1 L of water to make an isotonic solution.
To calculate the amount of NaCl needed to make an isotonic solution, we need to understand what an isotonic solution is.
An isotonic solution is a solution that has the same concentration of solutes (in this case, NaCl) as the fluid inside our cells. This means that when a cell is placed in an isotonic solution, there is no net movement of water into or out of the cell.
The average concentration of NaCl inside human cells is around 0.9% (w/v). This means that in 100 mL of solution, there should be 0.9 grams of NaCl. Since we want to prepare a 1 L solution, we need to scale up the amount of NaCl proportionally.
To calculate the amount of NaCl needed for 1 L of solution, we can use the following equation:
(0.9 grams / 100 mL) x (1000 mL / 1 L) = 9 grams
Therefore, to make an isotonic solution, you will need to add 9 grams of NaCl to 1 L of water.
You want a 0.9% solution of NaCl (by mass).
That is 0.9g/100 mL soln. So you want for 1000 mL just 0.9 g x (1000 mL soln/100 mL soln) = 9 g/L.