What volume of 1.20 M nitric acid is needed to convert 5.5 g of m- chlorophenol

Question

What volume of 1.20 M nitric acid is needed to convert 5.5 g of m- chlorophenol to 3-chloro-2, 6-dinitrophenol? (No excess of nitric acid is used.) In your experiment, you made quite a bit of another phenol. In addition to the one that you Isolated and purified. a. What is its structure? b. ON THE BASIS OF STRUCTURE, why did it stay behind during steam distillation? Explain C. ON THE BASS OF STRUCTURE, why did the compound you isolated distill over? What is the advantage of using steam distillation rather than simple distillation in this experiment? For each of the following. state whether it would undergo nitration faster, slower or at the same rate as phenol. Explain each answer. a. p-chlorophenol b. toluene c. p-nitrophenol

Explanation / Answer

(1)

m-Chlorophenol + 2 HNO3 ----------> 3-Chloro-2,6-dinitrophenol + 2 H2O

Mass of m-chlorophenol = 5.5 g.

Molar mass of m-chlorophenol = 128.56 g/mol

Number of moles of m-chlorophenol = mass / molar mass = 5.5 / 128.56 = 0.0428 mol

From the balanced equation,

1 mol of m-chlorophenol needs 2 mol of nitric acid

then, 0.0428 mol of m - chlorophenol needes 2*0.0428 = 0.0856 mol of nitric acid.

Thereofore,

Molarity = number of moles / volume of solution in L

1.20 = 0.0856 / V

V = 0.0856 / 1.20

V = volume of solution needed = 0.0713 L= 71.3 mL

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