Answer:
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Answer:
31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .
Explanation:
Given data:
Number of moles of CaCl₂ = 2.6 mol
Number of Cl₂ ions = ?
Solution:
CaCl₂ → Ca²⁺ + 2Cl⁻
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
In one mole of CaCl₂ there are two moles of chloride ions present.
In 2.6 mol:
2.6×2 = 5.2 moles
1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions
5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol
31.31× 10²³ number of Cl⁻
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
Primary producers i believe
