A chemist prepares a solution by dissolving 3.166 g of NaNO3 in enough water to make 200 mL of solution. What molar concentration of sodium nitrate should appear on the label? Answer in units of M.

 To calculate the molarity (M) of the solution, we need to follow these steps:

Step 1: Calculate the moles of NaNO3. We use the molar mass of NaNO3 to convert grams to moles. The molar mass of sodium nitrate (NaNO3) can be determined by adding the atomic masses of its constituent atoms: Na (sodium) = 22.99 g/mol, N (nitrogen) = 14.01 g/mol, O (oxygen) = 16.00 g/mol (and there are three oxygen atoms).

So, the molar mass of NaNO3 = 22.99 + 14.01 + (3 × 16.00) = 22.99 + 14.01 + 48.00 = 85.00 g/mol (rounded to two decimal places).

The number of moles of NaNO3 = mass (g) / molar mass (g/mol) The number of moles of NaNO3 = 3.166 g / 85.00 g/mol = 0.03724 mol (rounded to five significant figures).

Step 2: Determine the volume in liters. Since the molarity formula requires volume in liters, we convert milliliters to liters by dividing by 1000. 200 mL = 200 / 1000 L = 0.20 L

Step 3: Calculate the molarity. The molarity (M) is calculated by dividing the number of moles of solute by the volume of solution in liters. Molarity (M) = moles of solute / liters of solution Molarity (M) = 0.03724 mol / 0.20 L = 0.1862 M (rounded to four decimal places).

So, the chemist should label the solution as 0.1862 M sodium nitrate (NaNO3).