The Merck Index gives the following solubilities for benzoic acid in water: 0 C

Question

The Merck Index gives the following solubilities for benzoic acid in water:

0 C = 1.7 g/ L

20 C = 2.9 g/L

95 C = 68 g/L

If a chemist wants to crystallize 20 g of benzoic acid by using water as solvent at 95 °C, what would be the minimum amount of water required to dissolve it? If the chemist collected the crystals from the solution using the amount of water calculated in the prior question at 20 °C, how much benzoic acid would be obtained? If the chemist collected the crystals from the solution at 0 °C, how much benzoic acid would be obtained?

The answer from yahoo and from chegg are incorrect, only original answers please.

Explanation / Answer

Ans. Part 1: At 950C

Solubility of benzoic acid at 950C = 68 g/ 1000

That 1000 mL (= 1.0 L) of water dissolved in 68.0 g benzoic acid.

Amount of benzoic acid to be solvated = 20.0 g

Now,

Minimum volume of water required = Mass of benzoic acid / its solubility

                                                                        = 20.0 g / (68 g/ 1000 mL)

                                                                        = 294.12 mL = 0.29412 L

Note: It’s assumed that addition or removal of benzoic acid from solution does not affect its volume. That is, the volume of water remains constant at 294.12 mL.

Part 2: At 200C

We have, 294.12 mL solution containing 20.0 g benzoic acid- at 950C.

Maximum amount of benzoic acid that can be solvated in 294.12 mL of water at 200C is equal to-

                        Solubility at 200C x Volume of solution

                        = (2.9 g/ L) x (0.29412 L)

                        = 0.852948 g

Therefore, the specified volume of water can retain a maximum of 0.852948 g benzoic acid in solvated state at 200C, the rest amount would be precipitated out.

So,

            Amount of benzoic acid obtained (precipitated) = 20.0 g - 0.85 g

                                                                                                = 19.15 g

Part 3: At 00C

We have, 294.12 mL solution containing 20.0 g benzoic acid- at 950C.

Maximum amount of benzoic acid that can be solvated in 294.12 mL of water at 00C is equal to-

                        Solubility at 00C x Volume of solution

                        = (1.7 g/ L) x (0.29412 L)

                        = 0.50 g

Therefore, the specified volume of water can retain a maximum of 0.50 g benzoic acid in solvated state at 00C, the rest amount would be precipitated out.

So,

            Amount of benzoic acid obtained (precipitated) = 20.0 g - 0.50 g

                                                                                                = 19.50 g

Note: In part 3, the original solution is brought from 950C to 00C. Because it’s nowhere mentioned to cool solution 2 (the solution remaining after removal of benzoic acid at 200C) to 00C.

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