1. suppose that 3.00 g of naphthalene is dissolved in 30 mL of hot 95% ethanol a

Question

1. suppose that 3.00 g of naphthalene is dissolved in 30 mL of hot 95% ethanol and cooled to 10C. If the solubility of naphthalene is 1.20 g/100 mL of 95% ethanol at 10C, how much naphthalene remains unrecovered after filtration at 10C?

- the answer is 0.36 g, but I do not understand why the answer came out to be like this. I was thinking 0.36 g should be the amount that is recovered. Please explain.

2. How much more naphthalene is recoverable by evaporating the above 30 mL of mother liquor to 10 mL, again cooling to 10C and filtering?

Thank you in advance

Explanation / Answer

(1)

Solubility of naphthalene is given as, 1.20 g / 100 mL

So, 100 mL of 95% ethanol can dissolve 1.20 g. of naphthalene

then, 30 mL of 95% ethanol can dissolve 1.20 * 30 / 100 = 0.36 g.

So, 0.36 g. of naphthalene is in dissolved state which is not recovered.

So, recoverd amount = 3.00 - 0.36 = 2.64 g.

Unrecovered amount (or) amount that is present in dissolved state = 0.36 g.

(2) 10 mL of 95% ethanol can dissove 1.20 * 10 / 100 = 0.120 g.

Therefore, extra amount that can be recoverable = 0.36 - 0.12 = 0.24 g.

Because only 0.12 g. is in solution state. and 0.24 g. can be recovered.

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