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

Answer:

[NaNO₃] = 5.72 M

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

Answer 2

The concentration of the sodium nitrate solution has been 5.72 mol/L.

The concentration has been defined as the proportion of the solute to the solvent. The concentration with presence of 1 mole of solute in a liter of solution has been given by molarity.

Concentration of Sodium Nitrate

The molarity of the solution has been given by:

$\rm Molarity=\dfrac{Mass}{Molar\;mass}\;\times\;\dfrac{1000}{volume\;(mL)}$

The mass of sodium nitrate in the solution has been 122 g.

The molar mass of sample has been 85 g/mol

The volume of the solution has been 250 ml.

Substituting the values for Molarity of the solution as:

$\rm Molarity=\dfrac{122}{85}\;\times\;\dfrac{1000}{250}\\Molarity=1.435\;\times\;4\;mol/L\\Molarity=5.72\;g/mol$

The concentration of the sodium nitrate solution has been 5.72 mol/L.

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