A saline solution is 0.9% NaCl. What masses of NaCl and water would be required to prepare 50. L of this saline solution? Assume that the density of water is 1.000g/mL and that the NaCl does not add to the volume of the solution.
0.9% NaCl = 0.9 g/100 mL (assume density of 1.00 g/mL and NaCl doesn't add to the volume).
So that would be 0.9 x (50,000 mL/100 mL) = ??
Physiological saline solutions used in intravenous in-
jections have a concentration of 0.90% (mass/volume).
(a) How many grams of are needed to prepare
500.0 mL of this solution?
450 grams
Well, preparing a saline solution is as salty as figuring out your ex's password. Let's break down the math involved in making this solution a reality!
First, we need to find the mass of NaCl required. Since the saline solution is 0.9% NaCl, we know that 0.9% of the total mass of the solution will be NaCl.
To calculate the mass of NaCl, we can use the equation:
Mass of NaCl = (0.9% / 100%) x Total mass of the solution
Since we want to prepare 50 L of the saline solution, we'll convert it to grams since we're dealing with mass:
Total mass of the solution = 50 L x 1000 g/L = 50000 g
Now, let's solve for the mass of NaCl:
Mass of NaCl = (0.9% / 100%) x 50000 g = 450 g
So, you would need 450 g of NaCl to prepare 50 L of the saline solution.
Now, let's move on to the water. Since the NaCl does not add to the volume of the solution, we know that the remaining mass will be water.
Mass of water = Total mass of the solution - Mass of NaCl
Mass of water = 50000 g - 450 g = 49550 g
Therefore, you would need 49550 g (or 49.55 kg) of water to prepare 50 L of the saline solution.
Ahh, chemistry, turning boring ol' water into a salty solution!
To determine the mass of NaCl and water required to prepare the saline solution, we first need to find the mass of NaCl.
The concentration of the saline solution is given as 0.9% NaCl, which means that for every 100 ml of solution, there is 0.9 g of NaCl present.
To find the mass of NaCl required for 50 L (or 50,000 mL) of solution, we can set up a proportion.
(0.9 g NaCl / 100 mL) = (x g NaCl / 50,000 mL)
Cross-multiplying and solving for x, we get:
x = (0.9 g NaCl / 100 mL) * 50,000 mL
x = 450 g NaCl
Therefore, 450 grams of NaCl would be required to prepare 50 L of the saline solution.
Now, to find the mass of water required, we can subtract the mass of NaCl from the total mass of the solution.
Since the density of water is given as 1.000 g/mL, the mass of water can be found by multiplying the volume of water by the density.
Mass of water = (Volume of water) * (Density of water)
Mass of water = (50,000 mL - 0 mL) * (1.000 g/mL)
Mass of water = 50,000 g
Therefore, 50,000 grams of water would be required to prepare 50 L of the saline solution.