How much solid ammonium chloride would be needed to make 100 mL of a 0.5M solution?
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.
To calculate the amount of solid ammonium chloride needed to make a 0.5M solution, you will need to know the molar mass of ammonium chloride (NH4Cl).
The formula mass of a substance is the sum of the atomic weights of all the atoms in its empirical formula. In the case of ammonium chloride, the molar mass can be calculated as follows:
Molar mass of NH4Cl = (1 x molar mass of N) + (4 x molar mass of H) + (1 x molar mass of Cl)
Using the periodic table, the molar masses are:
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of Cl = 35.45 g/mol
So, the molar mass of NH4Cl = (1 x 14.01) + (4 x 1.01) + (1 x 35.45) = 53.49 g/mol
To prepare a 0.5M solution of ammonium chloride, we need to dissolve 0.5 moles of NH4Cl in 1 liter (1000 mL) of solution.
The amount of solid ammonium chloride needed can be calculated using the following formula:
Mass = (Molarity x Volume x Molar mass) / 1000
Mass = (0.5 mol/L x 100 mL x 53.49 g/mol) / 1000
Mass = (26.745 g)
Therefore, you would need approximately 26.745 grams of solid ammonium chloride to make a 0.5M solution in 100 mL.