How many grams of MgCl2 do you need to make 2 L of a 0.2 M solution?
To determine the number of grams of MgCl2 needed to make a 0.2 M solution in 2 L, you need to follow a step-by-step process. Here's how you can calculate it:
Step 1: Determine the molar mass of MgCl2.
The molar mass of magnesium chloride (MgCl2) can be calculated by adding up the atomic masses of each element. The atomic mass of magnesium (Mg) is 24.31 g/mol, and chlorine (Cl) has an atomic mass of 35.45 g/mol. Since there are two chlorine atoms in MgCl2, multiply the atomic mass of chlorine by 2.
Molar mass of MgCl2 = (1 * atomic mass of Mg) + (2 * atomic mass of Cl)
= (1 * 24.31 g/mol) + (2 * 35.45 g/mol) = 95.21 g/mol
Step 2: Calculate the number of moles required for a 0.2 M solution.
The molar concentration (M) is defined as the number of moles (n) of solute divided by the volume (V) of the solution in liters.
Molarity (M) = moles of solute (n) / volume of solution (V)
Since we want a 0.2 M solution and the volume is 2 L, we can rearrange the formula to solve for the number of moles (n) of MgCl2.
n = Molarity (M) * volume of solution (V)
= 0.2 mol/L * 2 L
= 0.4 moles
Step 3: Calculate the mass required for the calculated moles of MgCl2.
To determine the mass of MgCl2 needed in grams, we can use the molar mass and the number of moles.
Mass (g) = number of moles (n) * molar mass (g/mol)
= 0.4 moles * 95.21 g/mol
= 38.08 grams
Therefore, you would need approximately 38.08 grams of MgCl2 to make a 2 L 0.2 M solution.