How many grams of MgCl2 is needed to make 1200ml of a .25M solution?
.25M/L* 1200mL/1* 1L/1000mL* 95.211:-/ mol
I got 28.563g
Your answer is correct but you have too many significant figures listed. You're allowed only 2. That means to round your answer to 29 g.
To find the number of grams of MgCl2 needed to make a 0.25M solution in 1200ml, we need to use the formula:
grams = (molarity x volume x molar mass) / 1000
First, let's convert the volume from milliliters to liters:
1200ml = 1200 / 1000 = 1.2 liters
Next, we need to find the molar mass of MgCl2. The molar mass of MgCl2 can be calculated by adding the atomic masses of magnesium (Mg) and chlorine (Cl):
Mg: 24.31 g/mol
Cl: 35.45 g/mol
Molar mass of MgCl2 = (24.31 g/mol) + 2(35.45 g/mol) = 95.21 g/mol
Now, we can substitute these values into the formula:
grams = (0.25 mol/L x 1.2 L x 95.21 g/mol) / 1000
Calculating this expression gives us:
grams = (0.25 x 1.2 x 95.21) / 1000 = 28.563 grams
Therefore, you are correct. To make a 0.25M solution in 1200ml, you would need approximately 28.563 grams of MgCl2.