Answer:
To calculate the amount of ammonium chloride needed to make a 0.49 M solution in 5.2 L, we can use the formula:
Molarity (M) = moles of solute / liters of solution
First, we need to convert the volume of the solution from liters to milliliters (1 L = 1000 mL), so 5.2 L = 5200 mL.
Next, we rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) x liters of solution
moles of solute = 0.49 mol/L x 5.2 L = 2.548 mol
Since we know that the molar mass of ammonium chloride (NH4Cl) is approximately 53.49 g/mol, we can now calculate the mass of ammonium chloride needed:
mass = moles x molar mass
mass = 2.548 mol x 53.49 g/mol ≈ 136.38 grams
Therefore, approximately 136.38 grams of ammonium chloride would be required to make a 5.2 L solution with a concentration of 0.49 M.
Explanation:
