Question

If 4.49 g NaNO3 is dissolved in enough water to make 250.0 mL of solution, what is the molarity of the sodium nitrate solution?

Answer 1

Step-by-step explanation:

To find the molarity (M) of the sodium nitrate solution, we need to calculate the number of moles of NaNO3 and then divide it by the volume of the solution in liters.

First, let's calculate the number of moles of NaNO3 using its molar mass:

Molar mass of NaNO3 = Na (22.99 g/mol) + N (14.01 g/mol) + 3O (16.00 g/mol)

= 22.99 g/mol + 14.01 g/mol + (3 * 16.00 g/mol)

= 85.00 g/mol

Now, let's calculate the number of moles of NaNO3:

Number of moles = mass / molar mass

= 4.49 g / 85.00 g/mol

≈ 0.0529 mol

Next, we convert the volume from milliliters (mL) to liters (L):

Volume of solution = 250.0 mL = 250.0 mL × (1 L / 1000 mL)

= 0.250 L

Finally, we can calculate the molarity (M) using the formula:

Molarity (M) = number of moles of solute / volume of solution (in liters)

Molarity (M) = 0.0529 mol / 0.250 L

= 0.2116 M

Answer : 0.2116 M