# 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**.