A scientist makes an acid solution by adding drops of acid to 1.2 l of water. the final volume of the acid solution is 1.202 l.
1 answer:
Answer:- 40 drops and the percentage of acid in acid solution is 0.166%.
Solution:- Acid is added drops wise to 1.2 L of water to make acid solution. The final volume of the acid solution is 1.202 L. From here we could calculate the volume of the acid added to water by subtracting water volume from final acid solution volume as:
volume of acid added = 1.202 L - 1.2 L = 0.002 L
It's given that the volume of each drop is 0.05 mL. let's convert the volume of acid added from L to mL.
= 2mL
Now let's calculate the number of drops of acid added to water as:
= 40 drops
0.002 L of acid are present in 1.202 L of solution. It asks to calculate about what percent of acid solution is acid means how many L of acid are present in 100 L of acid solution. It's done as:
percentage of acid in acid solution =
= 0.166%
So, 40 drops of acid are required and there is 0.166% of acid in acid solution.
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The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.
Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).
Nomenclature:
In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.
Name of Formate:
Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.
E.g.
HCOOH (formic acid) → HCOO⁻ (formate) + H⁺
H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺
Formal Charges:
Formal charges are calculated using following formula,
F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]
For Oxygen:
F.C = [6] - [6 + 2/2]
F.C = [6] - [6 + 1]
F.C = 6 - 7
F.C = -1
For Sodium:
F.C = [1] - [0 + 0/2]
F.C = [1] - [0]
F.C = 1 - 0
F.C = +1
