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anygoal [31]
3 years ago
10

Suppose that there are two components, A and B, in a mixture. The chemical affinity of A for the stationary phase is less than t

hat of B and the chemical affinity of A for the mobile phase is greater than that of B. If the mixture is subjected to paper chromatography, which substance will have traveled further at the completion of the experiment?Which substance will have the largest Rf value?
Chemistry
1 answer:
kherson [118]3 years ago
8 0

Answer: Substance A will travel further at the end of the experiment and also it will have larger Rf value compared to substance B.

Explanation:

The affinity of substance towards the stationary or mobile phase means how much the substance binds to a phase. Because component B has greater affinity for the stationary phase, it binds more to stationary phase; while component A binds more to a mobile phase. Due to differences in their affinity, component A and B can be separated using chromatography. Stationary phase, as its name says, stays fixed, while the mobile phase is moving. Together with the mobile phase travels component A because of its greater affinity, while component B stays fixed due to its greater affinity for the stationary phase. In conclusion , substance A will travel further at the end of the experiment.

Rf value (Retention Factor) can be defined as the ratio of the distance traveled by a substance in the mobile phase (solute) compared with the distance traveled by the front of the mobile phase itself (solvent front).

Rf = Migration distance of substance / Migration distance of solvent front

Because the solvent front is always larger from the distance traveled by the solute, Rf values are always between 0 and 1. The farther a compound traveled, the larger its Rf value will be. In this example, substance A will have larger Rf value compared to substance B.

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<u>Answer:</u> The molarity of saline solution is 0.909 M

<u>Explanation:</u>

We are given:

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This means that 5.21 grams of NaCl is present in 100 grams of solution

To calculate mass of a substance, we use the equation:

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Density of solution = 1.02 g/mL

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Putting values in above equation, we get:

1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.02g/mL}=98.04mL

To calculate the moalrity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of NaCl = 5.21 g

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Putting values in above equation, we get:

\text{Molarity of solution}=\frac{5.21\times 1000}{58.44g/mol\times 98.04}\\\\\text{Molarity of solution}=0.909M

Hence, the molarity of saline solution is 0.909 M

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