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AP Chemistry 3.9 Separation of Solutions and Mixtures Chromatography- Exam Style questions - FRQs- New Syllabus

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Question

A student conducts a chromatography experiment and needs to prepare \( 100.0\ \mathrm{mL} \) of \( 0.340\ \mathrm{M\ NaCl(aq)} \) to use as the solvent.
(a) Calculate the mass of solid \( \mathrm{NaCl} \) (molar mass \( 58.44\ \mathrm{g/mol} \)) needed to prepare the \( 100.0\ \mathrm{mL} \) of \( 0.340\ \mathrm{M\ NaCl(aq)} \).
(b) In the following table, briefly list the additional steps necessary to prepare the \( 100.0\ \mathrm{mL} \) of \( 0.340\ \mathrm{M\ NaCl(aq)} \) solution using only materials selected from the choices given. Assume that all appropriate safety measures are already in place. Not all materials in the list may be needed.
• Solid \( \mathrm{NaCl} \)      • Distilled water      • Weighing paper and scoop
• Balance      • \( 100.0\ \mathrm{mL} \) volumetric flask      • \( 50.0\ \mathrm{mL} \) graduated cylinder
• Pipet      • \( 150\ \mathrm{mL} \) beakers      • Chromatography paper
StepStep Description and Materials Used
1.Use the weighing paper and scoop to measure the correct mass of solid \( \mathrm{NaCl} \) on the balance.
2. 
3.Swirl the mixture to dissolve the solid \( \mathrm{NaCl} \).
4. 
5.Stopper and invert the mixture several times to ensure that the mixture is homogeneous.
The student uses the \( \mathrm{NaCl(aq)} \) solvent to separate a mixture of compounds \( X \) and \( Y \) in a chromatography experiment. After \( 30\ \mathrm{minutes} \), the student removes the chromatography paper from the chamber. The results of the experiment are shown.
(c) A second student conducts the same chromatography experiment but removes the chromatography paper from the chamber after \( 15\ \mathrm{minutes} \) instead of \( 30\ \mathrm{minutes} \). Predict the effect, if any, this would have on the separation distance between compounds \( X \) and \( Y \) in the new experiment. Explain your reasoning.

Most-appropriate topic codes (AP Chemistry):

• Topic \( 3.7 \) — Solutions and Mixtures (Parts \( \mathrm{(a)} \), \( \mathrm{(b)} \))
• Topic \( 3.9 \) — Separation of Solutions and Mixtures (Part \( \mathrm{(c)} \))
▶️ Answer/Explanation

(a)
For the correct calculated value:
\( 0.1000\ \mathrm{L} \times \dfrac{0.340\ \mathrm{mol}}{1\ \mathrm{L}} \times \dfrac{58.44\ \mathrm{g}}{1\ \mathrm{mol}} = 1.99\ \mathrm{g\ NaCl} \)

First find the moles needed:
\( n = MV = (0.340)(0.1000) = 0.0340\ \mathrm{mol} \)
Then convert moles to grams using the molar mass.

(b)
For a correct description of step \( 2 \):
Combine the solid \( \mathrm{NaCl} \) and some distilled water in a \( 100.0\ \mathrm{mL} \) volumetric flask.

A little extra explanation: only add some water first so the solid can dissolve completely before the flask is filled to the final mark.

For a correct description of step \( 4 \):
Fill the volumetric flask with distilled water to the calibration \((100.0\ \mathrm{mL})\) mark.

The volumetric flask is used because it gives the correct final solution volume more accurately than a beaker or graduated cylinder.

(c)
For the correct prediction and a valid explanation:
It would decrease. The solvent front would not travel as far in the second experiment, so the separation would be smaller.

Because the paper is removed sooner, both compounds have less time to move with the solvent. Therefore, the distance between the \( X \) and \( Y \) spots would be reduced.

 
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