A Sodium Nitrate Solution: Calculating Molarity

A sodium nitrate solution is 15.6 % NaNO3 by mass and has a density of 1.02 g/mL. Calculate the molarity of the solution.

The molarity of the sodium nitrate solution is D) 1.87 M. To calculate the molarity of the solution, we first need to determine the mass of NaNO3 present in 1 L of the solution.

We can do this by multiplying the density (1.02 g/mL) by the volume (1000 mL) to get the mass of the solution, which is 1020 g/L.

Next, we need to calculate the mass of NaNO3 in 1 L of the solution. Since the solution is 15.6% NaNO3 by mass, we can multiply the mass of the solution (1020 g/L) by 0.156 to get the mass of NaNO3, which is 159.12 g/L.

Now, we can calculate the molarity of the solution using the formula:

Molarity = moles of solute / liters of solution

To convert the mass of NaNO3 to moles, we need to divide by its molar mass, which is 85.00 g/mol. Therefore, the number of moles of NaNO3 in 1 L of the solution is 159.12 g/L / 85.00 g/mol = 1.87 mol/L.

Therefore, the molarity of the sodium nitrate solution is D) 1.87 M.