When you performed a two-step dilution of the product for TLC, you weighed out o

Question

When you performed a two-step dilution of the product for TLC, you weighed out o.o5g of the product and 1.5 CH_2Cl_2. You then took 0.1 mL of the solution in a new test tube, and added 0.25 mL CH_1Cl_2 to make the working solution for TLC. If you only had 0.02 g of the product to begin with, how much CH_2Cl_2 would you need to make the same dilution (working solution) in one step? 0.50 mL 0.65 mL 1.5 mL 2.1 mL If the set of four compounds as shown below were subject to TLC, which one would have the lowest R_f in a non-polar mobile phase? The crude nitration product from nitration of methyl benzoate was isolated by vacuum Why filtration and washed with two portions of water, then two portions of ice-cold methanol. was methanol used to wash the crude product? It purifies it by removing the more methanol-soluble ortho and para products present To dissolve it for subsequent To neutralize the remaining sulfuric acid To cool the crude product and make it more crystalline

Explanation / Answer

1. 2 step dilution: First step: 0.05 g product + 1.5 mL CH2Cl2  = 0.05g /1.5 mL = 0.03 g/mL conc

Second Step: 0.1 mL above solution + 0.25 mL CH2Cl2 :

Since, 1mL of solution from step 1 = 0.03 g

Therefore, 0.1 mL = 0.1 * 0.03 g = 0.003 g

0.1 mL of solution in step contains 0.003 g of product

Adding 0.25 mL of CH2Cl2 in above solution:

Total volume of solution = 0.1+0.25 = 0.35 mL

0.35 mL soution contains 0.003 g product

1 mL solution = (1 mL/0.35 mL)*0.003 = 0.008 g

thus concentration of solution = 0.008 g/ mL

If the product is 0.02 g , amount of CH2Cl2 need to make the same dilution in one step:

Concentration of solution to be made = 0.008 g/ mL from 0.02 g product

0.008 g product require 1 mL of CH2Cl2

0.02 g product will require = (0.02 g / 0.008 g) * 1 mL = 2.5 mL

Thus 0.02 g product require 2.5 mL CH2Cl2 to make diltution so to obtain same concentration working solution.

2. Lowest Rf value would be of C. benzyl alcohol because it will participate in hydrogen bonding with stationary phase and would be less interacted with non polar mobile phase.

3. Washed with methanol because the ortho and para products are more soluble in methanol and impurities can be removed. Pure meta product can be obtained

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