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GarryVolchara [31]
3 years ago
6

What is the concentration (M) of CH3OH in a solution prepared by dissolving 34.4 g of CH3OH in sufficient water to give exactly

230 mL of solution
Chemistry
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

= 32g/mol

mole = 34.4/32

mole = 1.075mol

The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

Molarity = 4.6739

Molarity = 4.67M

The concentration of CH3OH in solution is 4.67M

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What amount of acid is esterified if a mixture consist initially of 1.0mole of ethanoic acid,1.0mole of ethanol and 1.0mole of w
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Answer:

0.54 mole

Explanation:

                                     CH3COOH    CH3CH2OH   CH3COOCH2CH3    H2O​

Initial concentration   1.0 mole         1.0 mole                0 mole                  1.0mol

Change                          - x                   - x                      + x                          + x  

Equilibrium                   (1.0 - x)          (1.0 - x)                   x                       (1.0 + x)

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x*(1.0+x)/(1.0-x)(1.0-x) = 4.0

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x+x² = 4(1² - 2x + x²)

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4 - 8x + 4x²- x² - x= 0

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2.5 mole of acid cannot be esterified, because there is only 1.0 mole of acid,

so answer is 0.54 mole.

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