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Fynjy0 [20]
3 years ago
9

Acetone major species present when dissolved in water

Chemistry
1 answer:
jek_recluse [69]3 years ago
3 0

Answer: acetone molecule ( CH₃-CO-CH₃)


Explanation:


1) Acetone is CH₃-CO-CH₃


2) That is a molecule (build up of covalent bonds).


3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.


This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.


That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).



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The density of an aqueous solution of nitric acid is 1.64 g/mL and the concentration is 1.85 M. What is the concentration of thi
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Mass % of the solution = 7.1067 %

Explanation:

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<u>Molarity of a solution is defined as the number of moles of solute present in 1 liter of the solution.</u>

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Lets, consider the volume of the solution = 1 L

Thus,

Moles of nitric acid present in the solution:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles of Nitric acid=Molarity \times {Volume\ of\ the\ solution}

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The mass of Nitric acid can be find out by using mole formula as:

moles=\frac{Mass\ taken}{Molar\ mass}

Thus,  

Mass\ of\ Nitric\ acid=Moles \times Molar mass}

Mass\ of\ Nitric\ acid=1.85 g \times 63 g/mol}

<u>Mass of Nitric acid = 116.55 g</u>

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Density=\frac{Mass}{Volume}

Given : Density = 1.64 g/mL

Also, 1 L = 10³ mL

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Mass\ of\ the\ solution=Density \times {Volume\ of\ the\ solution}

Mass\ of\ the\ solution=1.64 g/mL \times {1000 mL}

<u>Mass of the solution  = 1640 g</u>

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Mass \% =\frac{Mass\ of\ the\ solute}{Mass\ of\ the\ solution} \times {100}

So,

Mass \%=\frac{116.55}{1640} \times {100}

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