Question

The molar concentration (M) of a solution prepared by dissolving 0.2362g of Cr(NO3)3 in a 50-mL volumetric flask is 0.01985M, where the molecular weight for Cr(NO3)3 = 238.01g/mol.
a. Suppose you want to prepare another solution containing chromium nitrate that is 25 times LESS concentrated than the one prepared above. Given a choice of 10-mL and 5-mL pipets and 50-mL and 100-mL volumetric flasks, explain how you would proceed in preparing the new diluted solution. In addition, calculate the concentration for the new diluted solution. SHOW ALL WORK. Your final value should have the correct unit and number of significant figures. Hint: You will most likely need two dilution steps in order to obtain the desired concentration. Note: You may NOT reuse the same pipet or combine different pipets within the same dilution step. You may reuse the pipet and/or volumetric flask in the different dilution step.

Answer 1

Answer:

Here's what I get  

Explanation:

You want to dilute the original solution by a factor of 25 in two steps, so you could dilute it by a factor of 5 in the first step, then dilute the new solution by another factor of 5.

A. First dilution

Use a 10 mL pipet to transfer 10 mL of the original solution to a 50 mL volumetric flask. Make up to the mark with distilled water. Shake well to mix.

Use the dilution formula to calculate the new concentration.

$\begin{array}{rcl}c_{1}V_{1} & = & c_{2}V_{2}\\0.01985 \times 10.00 & = & c_{2} \times 50.00\\0.1985 & = & 50.00 c_{2}\\\\c_{2}& = & \dfrac{0.1985}{50.00}\\\\& = & \text{0.003 970 mol/L}\\\end{array}$

B. Second dilution

Repeat Step 1, using the 0.003 970 mol·L⁻¹ solution.

$\begin{array}{rcl}c_{2}V_{2} & = & c_{3}V_{3}\\0.003970 \times 10.00 & = & c_{3} \times 50.00\\0.03970 & = & 50.00 c_{3}\\\\c_{3}& = & \dfrac{0.03970}{50.00}\\\\& = & \textbf{0.000 7940 mol/L}\\\end{array}\\\text{The concentration of the final solution is \boxed{\textbf{0.000 7940 mol/L}} }$

3. Check:

Compare the final concentration with the original

$\begin{array}{rcl}\dfrac{ c_{3}}{ c_{1}} & = & \dfrac{0.0007940}{0.01985}\\& = & \mathbf{\dfrac{1}{25.00}}\\\end{array}\\\text{The concentration of the final solution is } \boxed{\mathbf{\dfrac{1}{25}}} \text{ that of the original solution}$