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Chemistry

1 answer:

Andru [333]3 years ago

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2 AL and 6 HCL AND 2 ALCL3 and 3 H2

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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, wh

Vinil7 [7]

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

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