Answer :

We have to calculate how much NaNO3 is needed to prepare a 225 mL of 1.55 M solution of NaNO3. Molarity (M) = (Amount of solute (in moles)) / (Volume of solution (in liters))

The answer to the given question is option A, which is 29.6 g.

Explanation: We have to calculate how much NaNO3 is needed to prepare a 225 mL of 1.55 M solution of NaNO3.

Molarity (M) = (Amount of solute (in moles)) / (Volume of solution (in liters))

We know, Amount of solute (in moles) = Molarity (M) × Volume of solution (in liters) = 1.55 M × 0.225 L = 0.34875 moles

We need to find the amount of NaNO3 in grams. For this, we need to use the following formula:

Amount of solute (in grams) = Amount of solute (in moles) × Molar mass of solute (in g/mol)

Molar mass of NaNO3 = (23 + 14 + 3×16) g/mol = 85 g/mol

Now, Amount of solute (in grams) = 0.34875 moles × 85 g/mol ≈ 29.6 g

Therefore, the amount of NaNO3 needed to prepare 225 mL of a 1.55 M solution of NaNO3 is approximately 29.6 g.

To know more about Molarity visit: https://brainly.com/question/2817451

#SPJ11

Thanks for taking the time to read How much NaNO3 is needed to prepare 225 mL of a 1 55 M solution of NaNO3 A 29 6 g B 0 244 g. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada