A chemist wants to make 4.5 L of a 0.310 M CaCl2 solution what mass of CaCl2 (in g) should the chemist use
How many moles do you want?
moles = M x L = ??
Then moles = grams/molar mass
Solve for grams.
Oh, chemistry! It's like mixing potions, but without the magic spells. To make your 0.310 M CaCl2 solution, let's do some calculations.
First, let's find the number of moles of CaCl2 needed. We can use the formula:
moles = concentration (M) x volume (L)
So, moles = 0.310 mol/L x 4.5 L = 1.395 moles of CaCl2.
Now, to find the mass of CaCl2 needed, we can use its molar mass. The molar mass of CaCl2 is approximately 110.98 g/mol.
So, mass (g) = moles x molar mass = 1.395 mol x 110.98 g/mol ≈ 154.27 g of CaCl2.
Voilà! To make 4.5 L of a 0.310 M CaCl2 solution, our chemist should use approximately 154.27 grams of CaCl2. But remember, always handle chemicals with care and never bring a clown to a chemistry lab!
To calculate the mass of CaCl2 required to make the solution, you need to consider the molar mass of CaCl2 and the desired concentration of the solution.
1. First, find the molar mass of CaCl2:
- Calcium (Ca) has an atomic mass of 40.08 g/mol.
- Chlorine (Cl) has an atomic mass of 35.45 g/mol.
- Since there are two chlorine atoms in CaCl2, the molar mass of CaCl2 is:
40.08 g/mol + (2 * 35.45 g/mol) = 110.98 g/mol.
2. Next, calculate the number of moles of CaCl2 required using the formula:
Moles = Concentration (M) * Volume (L)
Given:
Concentration = 0.310 M
Volume = 4.5 L
Moles = 0.310 M * 4.5 L = 1.395 moles
3. Finally, calculate the mass of CaCl2 using the formula:
Mass = Moles * Molar Mass
Mass = 1.395 moles * 110.98 g/mol = 154.36 g
Therefore, the chemist should use approximately 154.36 grams of CaCl2 to make a 4.5 L solution of 0.310 M CaCl2.