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wariber [46]
3 years ago
13

What is the molarity of formaldehyde in a solution containing 0.25 grams of formaldehyde per mL?

Chemistry
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

8.33mol/L

Explanation:

First, let us calculate the molar mass of of formaldehyde (CH2O). This is illustrated below:

Molar Mass of CH2O = 12 + (2x1) + 16 = 12 + 2 + 16 = 30g/mol

Mass of CH2O from the question = 0.25g

Number of mole CH2O =?

Number of mole = Mass /Molar Mass

Number of mole of CH2O = 0.25/30 = 8.33x10^-3mole

Now we can calculate the molarity of formaldehyde (CH2O) as follow:

Number of mole of CH2O = 8.33x10^-3mole

Volume = 1mL

Converting 1mL to L, we have:

1000mL = 1L

Therefore 1mL = 1/1000 = 1x10^-3L

Molarity =?

Molarity = mole /Volume

Molarity = 8.33x10^-3mole/1x10^-3L

Molarity = 8.33mol/L

Therefore, the molarity of formaldehyde (CH2O) is 8.33mol/L

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Answer:

The equilibrium constant is  K =0.02867

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Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

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    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

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