What is the mass of sodium nitrate NaNO3 that you would need to add to 250 g of water in order to prepare a solution that is 0.200 molal in NaNO3?
To determine the mass of sodium nitrate (NaNO3) needed to prepare a 0.200 molal solution, we need to understand the concept of molality and how it relates to the mass of the solute.
Molality (m) is a concentration unit that represents the number of moles of solute per kilogram of solvent. To calculate molality, we need to know the number of moles of solute and the mass of the solvent.
In this case, we want to prepare a 0.200 molal solution of NaNO3. This means that for every kilogram of water (solvent), we need 0.200 moles of NaNO3 (solute).
Step 1: Calculate the number of moles of NaNO3 needed.
To determine the number of moles, we can use the formula:
moles = molality × mass of solvent (in kg)
Given that the molality (m) is 0.200 and the mass of the solvent (water) is 250 g (0.250 kg), we can substitute these values into the formula:
moles = 0.200 × 0.250 = 0.050 mol
Step 2: Convert the moles of NaNO3 to grams.
To convert moles to grams, we need to know the molar mass of NaNO3, which is the sum of the atomic masses of its constituent elements: sodium (Na), nitrogen (N), and oxygen (O).
The molar mass of NaNO3 is:
(22.99 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 85.00 g/mol
Multiplying the moles of NaNO3 by its molar mass gives us the mass in grams:
mass = moles × molar mass
mass = 0.050 mol × 85.00 g/mol = 4.25 g
Therefore, you would need to add 4.25 grams of sodium nitrate (NaNO3) to 250 grams of water to prepare a solution that is 0.200 molal in NaNO3.