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Romashka [77]
3 years ago
14

A chemistry student is given 600. mL of a clear aqueous solution at 27.° C. He is told an unknown amount of a certain compound X

is dissolved in the solution. The student allows the solution to cool to 27.° C. The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 28.2 g.
Using only the information from above, can you calculate the solubility of X at 21.° C?
Chemistry
1 answer:
DiKsa [7]3 years ago
8 0

Answer: No

Explanation:

Firstly, the molar mass of the dissolved solid is not given. This is necessary to calculate the number of moles present in solution. Secondly, solubility always has to do with temperature and the specified temperature is 27°c and not 21°c. This makes it impossible to calculate the solubility at 21°c. Further information must supplied before the solubility at 21°c can be accurately calculated.

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The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

m=n\times M=0.7285 mol\times 236 g/mol=172 g

Therefore, mass of Al(C_{2}H_{3}O_{2})_{3} will be 172 g.

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