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Answer:
There are five <span>bonding pairs of electrons in Methanol.
Explanation:
Those electron pairs which are being shared between two atoms in molecule are called as bonding pair electrons. While, those electron pairs which are not involved in bond formation and are not shared between two atoms are called as Non-Bonding electron pairs.
In molecule of Methanol as shown below, it can be seen that carbon atom is forming four bonds with three hydrogen atoms and one oxygen atom by sharing four electron pairs and oxygen is forming two bonds, one with carbon atom and one with hydrogen atom. There are two lone pair of electrons present on oxygen atom which are not taking part in and formation.</span>
There are five <span>bonding pairs of electrons in Methanol.
Explanation:
Those electron pairs which are being shared between two atoms in molecule are called as bonding pair electrons. While, those electron pairs which are not involved in bond formation and are not shared between two atoms are called as Non-Bonding electron pairs.
In molecule of Methanol as shown below, it can be seen that carbon atom is forming four bonds with three hydrogen atoms and one oxygen atom by sharing four electron pairs and oxygen is forming two bonds, one with carbon atom and one with hydrogen atom. There are two lone pair of electrons present on oxygen atom which are not taking part in and formation.</span>
The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] = and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932