How many milliliters of a 0.50 M solution contain 14.0 g of BaCl2 (Barium Chloride)?
Molarity (M) = moles of solute/ volume of solution in liters
convert 14 g of BaCl2 to moles by dividing by the molar mass! this will give you the moles of solute.
once you plug every thing in and solve for volume, you would need to convert the liters into milliliters!
To determine the number of milliliters of a 0.50 M solution that contain 14.0 g of BaCl2, we first need to find the number of moles of BaCl2 and then use that information to calculate the volume.
1. Calculate the number of moles of BaCl2:
To do this, we need to know the molar mass of BaCl2. Barium (Ba) has a molar mass of 137.33 g/mol, while chloride (Cl) has a molar mass of 35.45 g/mol. Since BaCl2 has two chloride ions, we can calculate its molar mass as follows:
Molar mass of BaCl2 = (molar mass of Ba) + 2 x (molar mass of Cl)
= 137.33 g/mol + 2 x 35.45 g/mol
= 208.23 g/mol
Now we can calculate the number of moles of BaCl2:
Number of moles = mass of BaCl2 / molar mass of BaCl2
= 14.0 g / 208.23 g/mol
≈ 0.067 moles
2. Calculate the volume of the solution using the formula:
Volume (in liters) = Number of moles / Molarity
The molarity of the solution is given as 0.50 M, which means it contains 0.50 moles of BaCl2 per liter of solution. To convert this to milliliters, we need to multiply the volume in liters by 1000:
Volume (in milliliters) = Volume (in liters) x 1000
Substituting the values into the formula:
Volume (in liters) = 0.067 moles / 0.50 mol/L
= 0.134 L
Volume (in milliliters) = 0.134 L x 1000
= 134 mL
Therefore, 14.0 g of BaCl2 requires 134 mL of a 0.50 M BaCl2 solution.