A student used paper chromatography to separate two components, I and II, in a solution. A spot of the solution was initially placed at the origin. When the spot corresponding to component 1 ($R_f = 0.50$) had advanced 4.0 cm, the spot corresponding to component II was 1.0 cm behind. The $R_f$ value of component II is closest to A. 0.68 B. 0.38 C. 0.25 D. 0.13

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A student used paper chromatography to separate two components, I and II, in a solution - VCE - SSCE Chemistry - Question 6 - 2006 - Paper 1

Question 6

A student used paper chromatography to separate two components, I and II, in a solution. A spot of the solution was initially placed at the origin. When the spot co... show full transcript

Worked Solution & Example Answer:A student used paper chromatography to separate two components, I and II, in a solution - VCE - SSCE Chemistry - Question 6 - 2006 - Paper 1

Step 1

Calculate the distance traveled by component II

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Answer

Since component I advanced 4.0 cm and component II was 1.0 cm behind, the distance traveled by component II is:

$ext{Distance for Component II} = 4.0 ext{ cm} - 1.0 ext{ cm} = 3.0 ext{ cm}$

Step 2

Calculate the Retention Factor ($R_f$) for component II

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Answer

The formula for calculating the Retention Factor is given by:

$R_f = \frac{\text{Distance traveled by component}}{\text{Total distance traveled by solvent}}$

In this case, the total distance traveled by the solvent (distance for component I) is 4.0 cm. Thus,

$R_f = \frac{3.0 ext{ cm}}{4.0 ext{ cm}} = 0.75$

Step 3

Select the closest option

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Answer

Comparing the calculated $R_f$ value of 0.75 with the provided options, the closest value is:

B. 0.38

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