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tangare [24]
3 years ago
6

Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 2-butanol, C4H9OH; ma

gnesium chloride, MgCl2; and propane, C3H8. Rank from most to least soluble in water. To rank items as equivalent, overlap them.
Chemistry
1 answer:
inna [77]3 years ago
4 0

Answer:

MgCl2 > C4H9OH > CH4 > C3H8.

Explanation:

Alkanes do not form hydrogen bonds and are insoluble in polar solvents e.g water. The hydrogen bonds between water molecules are move away from an alkane molecule and this worsens as their Carbon chain / molecular weight increases.

MgCl2 is soluble in water. Water essentially breaks down the ionic crystal lattice and the resulting solution is slightly basic.

Alcohols are generally soluble in water and this is because of the -OH group and its ability to form hydrogen bonds with water molecules. As applied to alkanes, as the carbon chain in the alkyl group increases, the solubility decreases.

From the most soluble to the least soluble,

MgCl2 > C4H9OH > CH4 > C3H8.

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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
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Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

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