How many grams of (Bef2) are present in 655ml of a 0.442 M solution of (BeF2)
Please help
To determine the number of grams of BeF2 present in the given solution, you need to follow these steps:
Step 1: Calculate the number of moles of BeF2 using the molarity (M) and volume (in liters) of the solution.
Molarity (M) = moles of solute / liters of solution
Given:
Molarity (M) = 0.442 M
Volume of solution = 655 ml = 0.655 liters
moles of BeF2 = Molarity (M) × volume of solution (in liters)
moles of BeF2 = 0.442 M × 0.655 L
Step 2: Calculate the grams of BeF2 using the moles and molar mass of BeF2.
The molar mass of BeF2 = (beryllium (Be) atomic mass) + 2 × (fluorine (F) atomic mass)
The atomic masses can be found on the periodic table:
Be: 9.012 g/mol
F: 18.998 g/mol
Molar mass of BeF2 = 9.012 g/mol + 2 × 18.998 g/mol
grams of BeF2 = moles of BeF2 × molar mass of BeF2
Now, let's calculate the results:
moles of BeF2 = 0.442 M × 0.655 L
moles of BeF2 = 0.28961 mol
Molar mass of BeF2 = 9.012 g/mol + 2 × 18.998 g/mol
Molar mass of BeF2 = 47.998 g/mol
grams of BeF2 = 0.28961 mol × 47.998 g/mol
Therefore, there are approximately 13.89 grams of BeF2 present in 655 ml of a 0.442 M solution of BeF2.