How many grams of sodium chloride are needed to make 1 L of 0.9% (m/v) sodium chloride solution (“isotonic saline”).
0.9% m/v NaCl means 0.9g NaCl/100 mL.
You want to make a L of soln; therefore,
0.9g x (1000/100) = ?
Edith Cowan
To determine the grams of sodium chloride needed to make 1 L of a 0.9% (m/v) sodium chloride solution, you can follow these steps:
Step 1: Note that a 0.9% (m/v) solution means that there is 0.9 grams of solute (sodium chloride) dissolved in 100 mL of solution.
Step 2: Convert the volume from milliliters to liters. Since we want to make 1 L of the solution, the volume is 1 L.
Step 3: Use the proportion:
(0.9 g / 100 mL) = (x g / 1000 mL)
Solve for 'x' which represents the grams of sodium chloride needed.
Step 4: Cross-multiply and solve for 'x':
0.9 g * 1000 mL = 100 mL * 'x'
900 g = 100 'x'
Step 5: Divide both sides of the equation by 100:
900 g / 100 = 'x'
9 g = 'x'
So, you would need 9 grams of sodium chloride to make 1 L of a 0.9% (m/v) sodium chloride solution.
To determine the number of grams of sodium chloride needed to make 1 L of 0.9% sodium chloride solution, you need to follow these steps:
1. Understand the concept of "0.9% (m/v)" sodium chloride solution: "0.9% (m/v)" means that there are 0.9 grams of sodium chloride per 100 milliliters (0.9 g/100 mL) of solution.
2. Convert the given solution volume of 1 L into milliliters: There are 1000 milliliters in 1 liter, so we have 1000 mL.
3. Determine the amount of sodium chloride required for 0.9% concentration: Since the required concentration is 0.9 g/100 mL, we need to calculate the amount for 1000 mL.
- (0.9 g/100 mL) x (1000 mL/1 L) = 9 grams of sodium chloride
Therefore, to make 1 L of 0.9% (m/v) sodium chloride solution, you will need 9 grams of sodium chloride.