How many grams of CaCl2 would be required to be added to water to prepare 1 Liter (1000ml) of a 1.72% (w/v) aqueous solution of CaCl2?
1.72% w/v means 1.72 g/100 mL soln.
So if 1.72 grams are needed for 100 mL, then 17.2 g would be needed to prepare 1000 mL.
To calculate the number of grams of CaCl2 required to prepare a 1.72% (w/v) aqueous solution, you can follow these steps:
Step 1: Understand the given information.
The concentration of the solution is given as 1.72% (w/v), which means 1.72 grams of CaCl2 is present in 100 milliliters (ml) of solution.
Step 2: Define the desired final volume.
In this case, the desired final volume is 1 liter (1000 ml).
Step 3: Calculate the mass of CaCl2 required.
To find the mass of CaCl2 needed, we can set up a proportion using the given concentration:
(1.72 g CaCl2 / 100 ml solution) = (x g CaCl2 / 1000 ml solution)
Cross multiplying the proportion, we get:
x g CaCl2 = (1.72 g CaCl2 / 100 ml solution) * 1000 ml solution
Simplifying the equation, we have:
x g CaCl2 = (1.72 g CaCl2 / 100 ml) * 1000 ml solution
x g CaCl2 = 1.72 * 10 g CaCl2
Therefore, to prepare 1 liter (1000 ml) of a 1.72% (w/v) aqueous solution of CaCl2, you would need 17.2 grams (1.72 * 10 g) of CaCl2.