Respuesta :
[tex]\boxed{{\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}}[/tex] are present in 30 mL of 0.600 M [tex]{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}[/tex] solution.
Further Explanation:
The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:
1. Molarity (M)
2. Molality (m)
3. Mole fraction (X)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.
The formula to calculate the molarity of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex]] solution is as follows:
[tex]{\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}=\frac{{{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}{{{\text{Volume }}\left({\text{L}}\right){\text{ of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}}[/tex] …… (1)
Rearrange equation (1) to calculate the moles of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] solution.
[tex]\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}=\left({{\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\left({{\text{Volume of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\end{aligned}[/tex] …… (2)
The molarity of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] solution is 0.600 M.
The volume of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] solution is 30 mL.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}&=\left({{\text{0}}{\text{.600 M}}}\right)\left({{\text{30 mL}}}\right)\left({\frac{{{{10}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&=0.01{\text{8 mol}}\\\end{aligned}[/tex]
The dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] occurs as follows:
[tex]{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}\rightleftharpoons 2{\text{N}}{{\text{a}}^ + } + {\text{CO}}_3^{2 - }[/tex]
This indicates that one mole of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] dissociates to form two moles of [tex]{\text{N}}{{\text{a}}^ + }[/tex] ions and one mole of [tex]{\text{CO}}_3^{2 - }[/tex] ion. So total of three moles of ions are produced by one mole of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] .
The total number of moles of ions produced during the reaction is calculated as follows:
[tex]\begin{aligned}{\text{Moles of ions}}&=\left({{\text{0}}{\text{.018 mol}}}\right)\left({\text{3}} \right)\\&=0.0{\text{54 mol}}\\\end{aligned}[/tex]
The formula to calculate the number of ions is as follows:
[tex]{\text{Number of ions}}=\left({{\text{Moles of ions}}}\right)\left({{\text{Avogadro's Number}}}\right)[/tex] …… (3)
The number of moles of ions is 0.054 mol.
The value of Avogadro’s number is [tex]{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{ions}}[/tex] .
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Number of ions}}{\mathbf{ &= }}\left({0.054{\text{ mol}}}\right)\left( {\frac{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ ions}}}}{{{\text{1 mol}}}}} \right)\\&={\text{3}}{\text{.25188}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\&\approx {\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: molarity, ions, moles, Na2CO3, Na+, CO32-, 2Na+, 0.018 mol, 0.054 mol, Avogadro’s number, 3.25*10^22 ions, volume of Na2CO3, moles of Na2CO3.
The total number of ions present in 30.0 mL of 0.600M Na₂CO₃ solution is 3.25×10²² ions
We'll begin by calculating the number of mole of Na₂CO₃ in the solution. This can be obtained as follow:
Volume = 30 mL = 30 / 1000 = 0.03 L
Molarity = 0.6 M
Mole of Na₂CO₃ =?
Mole = Molarity x Volume
Mole of Na₂CO₃ = 0.6 × 0.03
Mole of Na₂CO₃ = 0.018 mole
- Next, we shall determine the total number of mole of ions in the solution.
Na₂CO₃(aq) —> 2Na⁺(aq) + CO₃²¯(aq)
From the balanced equation above,
1 mole of Na₂CO₃ contains 2 mole of Na⁺ and 1 mole of CO₃²¯.
Therefore,
0.018 mole of Na₂CO₃ will contain:
- 2 × 0.018 = 0.036 mole of Na⁺
- 0.018 mole of CO₃²¯
Mole of Na⁺ = 0.036 mole
Mole of CO₃²¯ = 0.018 mole
Total mole = 0.036 + 0.018
Total mole = 0.054 mole
- Finally, we shall determine the number of ions in the solution. This can be obtained as follow:
From Avogadro's hypothesis
1 mole = 6.02×10²³ ions
Therefore,
0.054 mole = 0.054 × 6.02×10²³
0.054 mole = 3.25×10²² ions
Therefore, the total number of ions present in the solution is 3.25×10²² ions
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