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Marina CMI [18]
4 years ago
14

For each of the following pairs of solutes and solvent, predict whether the solute would be soluble or insoluble. After making y

our predictions, you can check your answers by looking up the compounds in the Merck Index or the CRC Handbook of Chemistry and physics. Generally, the Merck Index I the easier reference book to use. If the substance has a solubility greater than 40mg/mL, you may conclude that it is soluble.
a.Malic acid in water: Soluble

b.Naphthalene in water: Insoluble

c.Amphetamine in ethyl alcohol: Insoluble

d.Aspirin in Water: Insoluble

e.Sucinic Acid in Hexane: Insoluble

f.Ibuprofen in Diethyl Ether: Insoluble

g.1-Decanol in Water: Slightly soluble because of OH
Chemistry
1 answer:
kakasveta [241]4 years ago
3 0

Answer and Explanation

The major rule of Solubility is that, 'like dissolves like', that is, polar solutes dissolve in polar solvents and non-polar solute dissolve in non-polar solvents. Polar solutes will not dissolve in non-polar solvents & vice-versa.

a) Malic Acid in Water - Soluble

Malic Acid, C₄H₆O₅, has a solubility of 558g/L in water at 25°C.

558 g/L = 558 mg/mL >> 40 mg/mL. This indicates that Malic Acid is very soluble in water.

Malic Acid is a dicarboxylic acid, therefore, it is a polar compound which is expected to be soluble in water as short chained polar organic compounds like itself are soluble in water.

b) Naphtalene in Water - Insoluble

Naphtalene, C₁₀H₈ has a solubility of 31.6 mg/L In water at 25°C.

31.6 mg/L = 0.0316 mg/mL <<< 40 mg/mL. This indicates that Naphtalene is very insoluble in water.

The insolubility of Naphtalene can be explained by the very non-polar nature of the organic compound.

c) Amphetamine in ethyl alcohol - Insoluble

Amphetamine, C₉H₁₃N has a solubility of 0.0165 g/mL in ethyl alcohol at 25°C.

0.0165 g/mL = 1.65 mg/mL << 40mg/mL

Amphetamine contains one benzene ring and one amine group. Even though, amine group makes the compound polar, the benzene ring and hydrocarbon chain overwhelm the polarity and cause amphetamine to be non-polar. Ethyl alcohol is polar due to having an alcohol functional group. By applying ‘Like dissolve like’, amphetamine is insoluble in ethyl alcohol.

d) Aspirin in water - Insoluble

Aspirin, C₉H₈O₄ has a solubility of 3mg/mL in water at 25°C.

3mg/mL << 40 mg/mL.

Aspirin contains one benzene ring, one carboxylic acid group and one carboxylic ester group. Even though, the carboxylic acid group and carboxylic ester group are polar, the benzene ring dominate and make aspirin nonpolar. Water is polar. By using ‘Like dissolve like’ rule, aspirin is insoluble in water.

e) Succinic acid in hexane - Insoluble.

Succinic acid, C₄H₆O₄ is insoluble in hexane.

Succinic acid contains two carboxylic acid groups which make the compound polar. However, hexane is nonpolar due to a long chain of hydrocarbon. By using ‘Like dissolve like’ rule, succinic acid is insoluble in hexane.

f) Ibuprofen in diethyl ether - Insoluble

Ibuprofen is insoluble in diethyl ether.

Ibuprofen contalns a complex chain of hydrocarbons with a benzene ring in between the chain and a carboxylic acid group. However, the big chain of hydrocarbons dominates the polarity

of the compound and makes it non-polar. Similarly, diethyl ether is a non-palar compound due the having an other group. By using 'Like dissolve like' rule, ibuprofen is soluble in diethyl ether since they are both nonpolar.

g) 1-Decanol (n-deryl alcohol) in water - slightly soluble.

1-decanol has a solubility of 37mg/L in water at 20°C.

37mg/L = 0.037 mg/mL << 40 mg/mL (Insoluble).

1-decanol is an alcohol. However, 1-decanol is a slightly polar compound since it has a 10-carbon chain and a hydroxyl group. Water is polar. So, because of this, 1-decanol is not so soluble in water.

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Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

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