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Chapter 4: Problem 54

Describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.707 M \mathrm{NaNO}_{3}\) solution.

Short Answer

Expert verified
To prepare 250 mL of a 0.707 M NaNO3 solution, you need approximately 15.02 grams of NaNO3.

Step by step solution

01

Convert the volume to liters

As the volume should be in liters when dealing with molarity, first convert the 250 mL to liters. We know that 1000 mL is equal to 1 L, so divide 250 by 1000 to convert mL to L: \(250 mL = 250 / 1000 L = 0.25 L\).
02

Calculate the amount of NaNO3 in moles

The molarity (M) of the solution is defined as moles of solute per liter of solution. We have a 0.707 M solution, meaning there are 0.707 moles of NaNO3 in one liter of solution. As we only have 0.25 L solution, we need to calculate the moles of NaNO3 needed for this volume: \(0.707 moles/L * 0.25 L = 0.17675 moles\).
03

Convert moles to grams

Now, convert the amount of NaNO3 from moles to grams to find out the weight of the NaNO3 required to make the solution. The molar mass of NaNO3 is 85 g/mol: \(0.17675 moles * 85 g/mol = 15.02 g\).

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