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Assoli18 [71]
3 years ago
9

How many carbon atoms are there in a diamond (pure carbon) with a mass of "48" mg?

Chemistry
1 answer:
blagie [28]3 years ago
7 0

Answer:

              3.47 × 10²³ C atoms

Solution:

Data Given:

                 Mass of Diamond  =  48 mg  =  0.048 g

                 M.Mass of Diamond  =  12.01 g.mol⁻¹ (as it is purely Carbon)

Step 1: Calculate Moles of Diamond as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  0.048 g ÷ 12.01 g.mol⁻¹

                 Moles  =  0.576 mol

Step 2: Calculate number of Carbon atoms,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Carbon atoms can be written as,

                 Moles  =  Number of C Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of C atoms,

                 Number of C atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting value of moles,

                 Number of C atoms  =  0.576 mol × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of C atoms =  3.47 × 10²³ C atoms

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The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

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Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

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<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

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