solution of 0.148 M sodium iodide for an experiment in lab, using a 500 mL volumetric flask. How much solid sodium iodide should you add
to do what? make 0.148 M NaI in 500 mL?
How many mols do you need? That's M x L = mols.
Then mols = grams/molar mass. You have mols and molar mass, solve for grams.
To determine the amount of solid sodium iodide you should add to prepare a 0.148 M solution in a 500 mL volumetric flask, you need to consider the definition of molarity and use some basic calculations.
Molarity (M) is defined as the number of moles of solute dissolved in one liter of solution. In this case, you want to prepare a 0.148 M sodium iodide solution in a 500 mL (0.5 L) volumetric flask.
The formula to calculate the number of moles is:
moles = Molarity × Volume (in liters)
Rearranging the formula, you can find the number of moles:
moles = Molarity × Volume (in liters)
= 0.148 M × 0.5 L
= 0.074 moles
Since sodium iodide (NaI) is a solid compound, you need to convert the number of moles to grams using the compound's molar mass. The molar mass of sodium iodide is 149.89 g/mol.
grams = moles × molar mass of NaI
= 0.074 mol × 149.89 g/mol
≈ 11.09 grams
Therefore, you should add approximately 11.09 grams of solid sodium iodide to the 500 mL volumetric flask to prepare a 0.148 M solution.