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GREYUIT [131]
3 years ago
10

What is the mass (in grams) of 9.79 × 1024 molecules of methanol (CH3OH)?

Chemistry
1 answer:
nevsk [136]3 years ago
6 0
We will use this formlula: Mass in grams = Number of moles x Molecular mass of 1 mole.

Since, we know the avagadro number is 6.02 x 10²³, we only have two unknown values left which are the molecular mass of CH3OH and its mole.

Molecular Mass: C = 12, H= 1, O = 16, since we have C=12, H4 = 4, O = 16, we will add them up: 12 + 4 + 16 =32

We know that one mole of anything = 6.02 x 10²³.
So we will use this formula to find the mole of methanol: Number of moles = Number of molecules / Avagadro number

Number of moles of CH3OH = (9.79 x 10^24)/6.02 x 10²³) =  16.263 moles.

Now we know that the molecular mass = 32 and the mole is = 16.263.

Now we can find its mass by using this formula: <span>Mass in grams = Number of moles x Molecular mass of 1 mole.
</span>
Mass in grams = 16.263 x 32 = 520g






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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

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

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

b=\frac{L}{mol}

T=K

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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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