How would you prepare 1 L of a 1.15 M aqueous solution of MgCl2? What grams of MgCl2 would be used?
You would weigh out exactly 1.5 moles of MgCl2 (molar mass MgCl2 x 1.5), place in a 1 L volumetric flask, add some water, swirl until all of the MgCl2 is dissolved, make to the mark of the volumetric flask with distilled water, mix thoroughly.
To prepare a 1.15 M aqueous solution of MgCl2, you would need to follow a few steps. Here's how you can do it:
Step 1: Determine the molar mass of MgCl2
The molar mass of magnesium chloride (MgCl2) can be calculated by adding the atomic masses of its individual elements. The atomic mass of magnesium (Mg) is 24.31 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol. Therefore, the molar mass of MgCl2 is:
MgCl2 = (1 * Mg) + (2 * Cl) = (1 * 24.31 g/mol) + (2 * 35.45 g/mol) = 95.21 g/mol.
Step 2: Calculate the amount (in moles) of MgCl2 needed
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. In this case, you want to prepare a 1.15 M solution of MgCl2 with a volume of 1 L. Therefore, you would need 1.15 moles of MgCl2.
Step 3: Convert moles to grams
To convert moles to grams, you can use the molar mass of MgCl2 calculated in step 1. Multiply the moles of MgCl2 by its molar mass to get the grams required:
grams of MgCl2 = moles of MgCl2 * molar mass of MgCl2
grams of MgCl2 = 1.15 moles * 95.21 g/mol = 109.4935 grams (rounded to four decimal places).
So, to prepare 1 L of a 1.15 M aqueous solution of MgCl2, you would need 109.4935 grams of MgCl2.