Question

A Chemist prepares a solution of sodium nitrate (NaNO3) by measuring out 122.g of sodium nitrate into 250.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium nitrate solution. Round your answer to 3 significant digits.

Answer 1

The concentration of the sodium nitrate solution prepared by the chemist is 5.74 mol/L, after calculating the number of moles of sodium nitrate and dividing it by the solution volume in liters.

Step-by-step explanation:

To calculate the concentration in mol/L (molarity) of the chemist's sodium nitrate solution, we first need to find the number of moles of sodium nitrate. This can be done by dividing the mass of the sodium nitrate by its molar mass. The molar mass of sodium nitrate (NaNO3) is approximately 85.00 g/mol.

Number of moles (n) = mass (g) / molar mass (g/mol) = 122 g / 85.00 g/mol = 1.435 mol (rounded to four significant figures)

We know the volume of the solution is 250 mL, which is 0.250 L. The concentration (C) of the solution in mol/L can then be calculated using:

Concentration (C) = number of moles (n) / volume (V) in liters = 1.435 mol / 0.250 L = 5.74 mol/L (rounded to three significant digits)

Therefore, the concentration of the sodium nitrate solution prepared by the chemist is 5.74 mol/L.

Answer 2

[NaNO₃] = 5.72 M

Step-by-step explanation:

Molarity is a sort of concentration.

It says the moles of solute, that are contained in 1L of solution.

The volume is in mL, so let's convert to L.

250 mL . 1 L / 1000 mL = 0.250 L

Let's convert now, the mass of NaNO₃, our solute into moles

122 g / 85 g/mol = 1.43 moles

Molarity = mol / L → 1.43 mol / 0.250L = 5.72 M