Which products are formed when aluminum is added to a silver chloride solution?
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The concentration of each ion in the solution of ammonium chloride is:
- NH₄⁺ = 0.1886 M
- Cl⁻ = 0.1886 M
To solve this problem, the formulas and the procedures that we have to use are:
- M = n(solute)/v(solution) L
- n = m / MW
- MW= ∑ AWT
Where:
- M= molarity
- n = moles
- m = mass
- v = volume
- MW = molecular weight
- AWT = atomic weight
Information about the problem:
- m(NH₄Cl) = 5 g
- v(solution) = 500 ml
- AWT (N)= 14 g/mol
- AWT (H)= 1 g/mol
- AWT (Cl) = 35 g/mol
Converting the volume units from (ml) to (L) we have:
v(solution) = 500 ml * (1 L/1000 ml)
v(solution) = 0,50 L
We calculate the moles of the NH₄Cl from the MW:
MW = ∑ AWT
MW (NH₄Cl)= AWT (N) + AWT (H)*4 + AWT (Cl)
MW (NH₄Cl)= 14 g/mol +( 1 g/mol * 4) + 35 g/mol
MW (NH₄Cl)= 14 g/mol + 4 g/mol + 35 g/mol
MW (NH₄Cl)= 53 g/mol
Having the MW we calculate the moles of NH₄Cl:
n(NH₄Cl) = m(NH₄Cl) / MW(NH₄Cl)
n(NH₄Cl) = m(H2SO4) / MW (H2SO4)
n(NH₄Cl) = 5 g / 53 g/mol
n(NH₄Cl) = 0.0943 mol
Applying the molarity formula, we get:
M(NH₄Cl) = n(NH₄Cl)/v(solution) L
M(NH₄Cl) = 0.0943 mol / 0,50 L
M(NH₄Cl) = 0.1886 M
There are 0.1886 moles of NH₄Cl per liter of solution.
Let's recognize that 1 mol NH₄Cl contains:
- 1 mol NH₄⁺
- 1 mol Cl⁻
The concentration of each ion is thus:
(1)*(0.1886 M) = 0.1886 M NH₄⁺
(1)*(0.1886 M) = 0.1886 M Cl⁻
<h3>What is a solution?</h3>In chemistry a solution is known as a homogeneous mixture of two or more components called:
- Solvent
- Solute
Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161
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