Copyright

Arrange the following from lowest to highest number of particles per L of solution: 0.1 M...

Question:

Arrange the following from lowest to highest number of particles per L of solution: {eq}\rm 0.1\ M\ Na_3PO_4,\ 0.1\ M\ C_2H_5OH,\ 0.01\ M\ CO_2,\ 0.15\ M\ NaCl,\ 0.2\ M\ CaCl_2 {/eq}.

Stoichiometry:

Stoichiometry can be understood as the proportion of a substance that actively takes part in a particular reaction. It is helpful in determining the quantity of a substance that would be required for any reaction.

Answer and Explanation: 1


Given Data:

  • A solution of {eq}{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} of molarity {eq}0.1\;{\rm{M}} {/eq}.
  • A solution of {eq}{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} {/eq}of molarity {eq}0.1\;{\rm{M}} {/eq}.
  • A solution of {eq}{\rm{C}}{{\rm{O}}_2} {/eq}of molarity {eq}0.01\;{\rm{M}} {/eq}.
  • A solution of {eq}{\rm{NaCl}} {/eq} of molarity {eq}0.15\;{\rm{M}} {/eq}.
  • A solution of {eq}{\rm{CaC}}{{\rm{l}}_2} {/eq} of molarity {eq}0.2\;{\rm{M}} {/eq}.
  • The volume of solution is {eq}1\;{\rm{L}} {/eq}


{eq}{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} {/eq}

The total number of moles in {eq}0.1\;{\rm{M}}\;{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} solution can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = {\rm{Molarity}} \times {\rm{Volume}}\;{\rm{in}}\;{\rm{L}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= 0.1\;{\rm{M}} \times 1\;{\rm{L}}\\ &= {\rm{0}}{\rm{.1}}\;{\rm{mol}} \end{align*} {/eq}


Number of particles of {eq}{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} in {eq}0.1\;{\rm{moles}} {/eq} can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{particles}} = {N_{\rm{A}}} \times {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{particles}} &= 6.022 \times {10^{23}} \times {\rm{0}}{\rm{.1}}\;{\rm{mol}}\\ &= 6.022 \times {10^{22}} \end{align*} {/eq}


Therefore, total number of particles in {eq}0.1\;{\rm{M}}\;{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} is {eq}{\rm{6}}{\rm{.022}} \times {\rm{1}}{{\rm{0}}^{22}} {/eq}.

{eq}{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} {/eq}

The total number of moles in {eq}0.1\;{\rm{M}}\;{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} {/eq} solution can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = {\rm{Molarity}} \times {\rm{Volume}}\;{\rm{in}}\;{\rm{L}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= 0.1\;{\rm{M}} \times 1\;{\rm{L}}\\ &= {\rm{0}}{\rm{.1}}\;{\rm{mol}} \end{align*} {/eq}


Number of particles of {eq}{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} {/eq} in {eq}0.1\;{\rm{moles}} {/eq} can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{particles}} = {N_{\rm{A}}} \times {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{particles}} &= 6.022 \times {10^{23}} \times {\rm{0}}{\rm{.1}}\;{\rm{mol}}\\ &= 6.022 \times {10^{22}} \end{align*} {/eq}


Therefore, total number of particles in {eq}0.1\;{\rm{M}}\;{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} {/eq}is {eq}{\rm{6}}{\rm{.022}} \times {\rm{1}}{{\rm{0}}^{22}} {/eq}.


{eq}{\rm{C}}{{\rm{O}}_2} {/eq}

The total number of moles in {eq}0.01\;{\rm{M}}\;{\rm{C}}{{\rm{O}}_2} {/eq} solution can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = {\rm{Molarity}} \times {\rm{Volume}}\;{\rm{in}}\;{\rm{L}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= 0.01\;{\rm{M}} \times 1\;{\rm{L}}\\ &= 0.01\;{\rm{mol}} \end{align*} {/eq}


Therefore, total number of particles in {eq}0.01\;{\rm{M}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq} is {eq}{\rm{6}}{\rm{.022}} \times {\rm{1}}{{\rm{0}}^{21}} {/eq}.


{eq}{\rm{NaCl}} {/eq}

The total number of moles in {eq}0.15\;{\rm{M}}\;{\rm{NaCl}} {/eq} solution can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = {\rm{Molarity}} \times {\rm{Volume}}\;{\rm{in}}\;{\rm{L}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} &= 0.15\;{\rm{M}} \times 1\;{\rm{L}}\\ &= 0.15\;{\rm{mol}} \end{align*} {/eq}


Number of particles of {eq}{\rm{NaCl}} {/eq} in {eq}0.15\;{\rm{moles}} {/eq} can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{particles}} = {N_{\rm{A}}} \times {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{particles}} &= 6.022 \times {10^{23}} \times {\rm{0}}{\rm{.15}}\;{\rm{moles}}\\ &= 9.033 \times {10^{22}} \end{align*} {/eq}


Therefore, total number of particles in {eq}0.15\;{\rm{M}}\;{\rm{NaCl}} {/eq} is {eq}{\rm{9}}{\rm{.033}} \times {\rm{1}}{{\rm{0}}^{22}} {/eq}.


{eq}{\rm{CaC}}{{\rm{l}}_2} {/eq}

The total number of moles in {eq}0.2\;{\rm{M}}\;{\rm{CaC}}{{\rm{l}}_2} {/eq} solution can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = {\rm{Molarity}} \times {\rm{Volume}}\;{\rm{in}}\;{\rm{L}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{moles}}&= 0.2\;{\rm{M}} \times 1\;{\rm{L}}\\ &= 0.2\;{\rm{mol}} \end{align*} {/eq}


Number of particles of {eq}{\rm{CaC}}{{\rm{l}}_2} {/eq} in {eq}0.2\;{\rm{moles}} {/eq} can be calculated by the given formula.

{eq}{\rm{Number}}\;{\rm{of}}\;{\rm{particles}} = {N_{\rm{A}}} \times {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} {/eq}


Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Number}}\;{\rm{of}}\;{\rm{particles}} &= 6.022 \times {10^{23}} \times {\rm{0}}{\rm{.2}}\;{\rm{mol}}\\ &= 1.2044 \times {10^{23}} \end{align*} {/eq}


Therefore, total number of particles in {eq}0.2\;{\rm{M}}\;{\rm{CaC}}{{\rm{l}}_2} {/eq} is {eq}{\rm{1}}{\rm{.2044}} \times {\rm{1}}{{\rm{0}}^{23}} {/eq}.


Therefore increasing order of number of particles per liter for the given solutions is shown below.

{eq}0.01\;{\rm{M}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}} < 0.1\;{\rm{M}}\;{\rm{N}}{{\rm{a}}_3}{\rm{P}}{{\rm{O}}_4} = 0.1\;{\rm{M}}\;{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}} < 0.15\;{\rm{M}}\;{\rm{NaCl}} < 0.2\;{\rm{M}}\;{\rm{CaC}}{{\rm{l}}_2} {/eq}


Learn more about this topic:

Loading...
Stoichiometry: Calculating Relative Quantities in a Gas or Solution

from

Chapter 9 / Lesson 4
24K

Learn how to find the molarity of a solution or the moles of gas in a given volume using stoichiometry. See examples of calculating moles from molarity.


Related to this Question

Explore our homework questions and answers library