You are asked to prepare \(125.0 \mathrm{mL}\) of \(0.0321 \mathrm{M} \mathrm{AgNO}_{3}\) How many grams would you need of a sample known to be \(99.81 \% \mathrm{AgNO}_{3}\) by mass?

Short Answer

Expert verified
You would need approximately 0.6823 grams of the \(99.81\% \,AgNO3\) sample.

Step by step solution

01

Understand the Question and Identify the Given Information

The question provides the following information:1. Volume (V) of the AgNO3 solution to be prepared = 125.0 mL or 0.125 L (Since one liter (L) is equivalent to 1000 mL).2. Molarity (M) of the AgNO3 solution to prepare = 0.0321 M.3. Purity of the AgNO3 sample = 99.81%.
02

Calculate the Moles of AgNO3 Needed

Using the molarity formula \(M = \frac{n}{V}\), where n represents number of moles and V is the volume, we can calculate the moles of AgNO3 needed for the solution. Rearranging the formula gives \(n = M \times V\), so \(n = 0.0321 M \times 0.125 L = 0.0040125 mol\).
03

Convert Moles to Grams

Using the molar mass of AgNO3 (169.87 g/mol), calculated from the sum of the molar masses of its constituent elements, the moles of AgNO3 are converted to grams with the formula \(mass = n \times molar \,mass\). This gives \(0.0040125 mol \times 169.87 g/mol = 0.6813 g\). However, this represents the mass needed if the sample was 100% pure.
04

Adjusting for the Purity of the Sample

Since the sample is only 99.81% pure, the mass calculated above represents only 99.81% of the total sample we need. Therefore, the total sample needed can be calculated as \(\frac{0.6813 g}{0.9981} = 0.6823 g\), adjusting for the purity of the sample.

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