how many grams of CaCl2 would be required to be added to water to prepare 1 liter (1000 ml) of a 1.72% (w/v) aqueous solution of CaCl2?
1.72% w/v means 1.72 g CaCl2 in 100 mL of solution so that would be 10x that for 1 L of solution or 17.2 g CaCl2. The answer to the problem, therefore, is 17.2 g CaCl2 but it makes a difference how the solution is prepared. You can NOT add 17.2 g CaCl2 to a liter (1000 mL) of water because that will be more than 1000 mL OF SOLUTION. You prepare it correctly by adding 17.2 g CaCl2 to a 1000 mL volumetric flask, adding some water, swirl until all of the CaCl2 is dissolved, then make the final volume to 1000 mL (by adding water to the mark on the flask).
To determine the number of grams of CaCl2 needed to prepare a 1.72% (w/v) aqueous solution in 1 liter (1000 ml) of water, we need to follow these steps:
Step 1: Convert the percentage concentration to grams.
In a 1.72% (w/v) solution, the percentage represents the weight of the solute (CaCl2) per volume of the solution (mL).
1.72% (w/v) means there are 1.72 grams of CaCl2 present in 100 ml of solution.
Step 2: Calculate the amount of CaCl2 required for the desired volume.
We know that we need to prepare a 1-liter solution (1000 ml). Since we have the concentration of CaCl2 in a 100 ml solution, we can use this ratio to find the mass required for 1000 ml.
(1.72 grams CaCl2 / 100 ml solution) = (X grams CaCl2 / 1000 ml solution)
Solving this ratio equation for X, we find:
X = (1.72 grams CaCl2 / 100 ml solution) * 1000 ml solution
X = 17.2 grams CaCl2
Therefore, you would need to add 17.2 grams of CaCl2 to water to prepare 1 liter (1000 ml) of a 1.72% (w/v) aqueous solution of CaCl2.