Calculate the grams of solute required to make 2500mL of 90% saline solution?
Surely you didn't mean 90% but that's what I read (and I assume that is 90% w/v). A 90% saline means 90 g NaCl/100 mL and you want 2500 mL; that will be
90g x (2500/100) = ? g NaCl.
To calculate the grams of solute required to make a 90% saline solution, we need to know the desired percentage concentration of the solute in the solution and the volume of the solution we want to make.
Given:
- Desired percentage concentration: 90%
- Volume of the solution: 2500 mL
To calculate the grams of solute required, we can use the formula:
grams of solute = (desired concentration / 100) * volume of solution
Let's substitute the given values into the formula:
grams of solute = (90 / 100) * 2500 mL
grams of solute = 0.9 * 2500 mL
grams of solute = 2250 grams
Therefore, to make 2500 mL of a 90% saline solution, you would need 2250 grams of solute.
To calculate the grams of solute required to make a saline solution, we need to know the concentration (as a percentage) and the volume of the solution.
In this case, the concentration is given as 90%, and the volume of the solution is 2500 mL.
Step 1: Convert the percentage to a decimal.
To convert 90% to decimal form, divide it by 100:
90% = 90/100 = 0.9
Step 2: Calculate the grams of solute using the formula:
Grams of Solute = Concentration (as a decimal) x Volume of Solution (in mL)
Grams of Solute = 0.9 x 2500 mL
Step 3: Simplify the equation by converting mL to grams (assuming density is 1 g/mL for simplicity).
Grams of Solute = 0.9 x 2500 g
Grams of Solute = 2250 g
Therefore, to make a 2500 mL saline solution with a concentration of 90%, you would need 2250 grams of solute.