How many liters of a 0.80%\% (m/v{\rm m/v}) NaCl{\rm NaCl} solution can be prepared from 500mLmL of a 9.0%\% (m/v{\rm m/v}) stock solution?
L1 x %1 = L2 x %2
?L x 0.8% = 0.5L x 9%
?L = ?
To solve this problem, we can use the concept of the concentration of a solution.
The given information tells us that we have a 9.0% (m/v) NaCl stock solution and we want to know how many liters of a 0.80% (m/v) NaCl solution we can prepare.
The general formula for calculating the concentration of a solution is:
C1V1 = C2V2
Where:
C1 = concentration of the initial solution (9.0%)
V1 = volume of the initial solution (500 mL)
C2 = concentration of the final solution (0.80%)
V2 = volume of the final solution (what we need to find)
Let's rearrange the formula to solve for V2:
V2 = (C1V1) / C2
Now, let's substitute the given values into the formula and solve for V2:
V2 = (9.0% * 500 mL) / 0.80%
First, we convert the percentages to decimal values by dividing by 100:
V2 = (0.09 * 500 mL) / 0.008
Next, we perform the multiplication:
V2 = 4.5 mL / 0.008
Now, we divide:
V2 = 562.5 mL
Finally, since the question asks for the volume in liters, we convert the mL to liters by dividing by 1000:
V2 = 562.5 mL / 1000 mL/L
V2 = 0.5625 L
Therefore, we can prepare approximately 0.5625 liters (or 562.5 mL) of a 0.80% (m/v) NaCl solution from 500 mL of a 9.0% (m/v) NaCl stock solution.