What is the mass of Na2SO4 is needed to make 1.5 L of 3.0 M solution? Na=23 g.: s = 32 g: O=16g.)
To find the mass of Na2SO4 needed to make a 1.5 L solution of 3.0 M concentration, you need to follow a few steps. Here's how:
Step 1: Determine the molar mass of Na2SO4.
The molar mass of Na2SO4 can be calculated by adding up the atomic masses of each constituent element:
Na (sodium): 2 atoms x 23 g/mol = 46 g/mol
S (sulfur): 1 atom x 32 g/mol = 32 g/mol
O (oxygen): 4 atoms x 16 g/mol = 64 g/mol
Adding these together, we get:
Molar mass of Na2SO4 = 46 g/mol + 32 g/mol + 64 g/mol = 142 g/mol
Step 2: Calculate the moles of Na2SO4 needed.
Using the given concentration (3.0 M) and volume (1.5 L), we can apply the formula:
Molarity (M) = Moles (mol) / Volume (L)
Rearranging the formula, we have:
Moles (mol) = Molarity (M) x Volume (L)
Plugging in the values:
Moles (mol) = 3.0 M x 1.5 L = 4.5 mol
Step 3: Calculate the mass of Na2SO4 needed.
To convert moles to grams, we can multiply the moles by the molar mass of Na2SO4:
Mass (g) = Moles (mol) x Molar mass (g/mol)
Plugging in the values:
Mass (g) = 4.5 mol x 142 g/mol = 639 g
Therefore, you will need 639 grams of Na2SO4 to make a 1.5 L solution with a concentration of 3.0 M.
How many mols do you need?
That's mols = M x L = ?
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.