Chapter 2, Problem 2.72QA

Interpretation Introduction

To find:

The numbers of moles of Na­­­⁺ ions are there in 1 mole of the given compounds.

Answer to Problem 2.72QA

Solution:

a) $1$ mol of Na_­­­⁺.

b) $2$ mol of Na_­­­⁺.

c) $3$ mol of Na_­­­⁺.

d) $1$ mol of Na_­­­⁺.

Explanation of Solution

Calculations:

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

a) $1$ mole of NaCl

We can write the dissociation of NaCl from Na⁺ and Cl^- as follows,

$\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{ }\to \mathrm{ }{\mathrm{N}\mathrm{a}}^{+}+\mathrm{ }{\mathrm{C}\mathrm{l}}^{-}$

As from the equation we can easily determine the mole to mole ratio as here the mole ratio between NaCl and Na⁺ is $1:1$.

So, now we have the mole to mole ratio and moles of NaCl, so we can calculate the moles of Na⁺ by using dimensional analysis as below,

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{*}\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}}=1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}$

So, $1$ mole of NaCl contains $1$ mole of Na⁺.

b) $1$ mole of Na₂SO₄

The dissociation of Na₂SO₄ from Na⁺ and SO₄^2- can be given as below;

${\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4\mathrm{ }\to \mathrm{ }{2\mathrm{N}\mathrm{a}}^{+}+\mathrm{ }{{\mathrm{S}\mathrm{O}}_4}^{2-}$

From the given balanced equation, the mole to mole ratio between Na₂SO₄ and Na⁺ is $1:2$.

So, by using dimensional analysis, we can calculate the moles of Na⁺ present in 1 mol of Na₂SO₄ as follows:

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4\mathrm{*}\frac{2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}_2{\mathrm{S}\mathrm{O}}_4}=2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}$

So, $1$ mole of Na₂SO₄ contains $2$ mole of Na⁺.

c) $1$ mole of Na₃PO₄

The dissociation of Na₃PO₄ from Fe and OH^- can be given as below

${\mathrm{N}\mathrm{a}}_3{\mathrm{P}\mathrm{O}}_4\mathrm{ }\to \mathrm{ }{3\mathrm{N}\mathrm{a}}^{+}+\mathrm{ }{{\mathrm{P}\mathrm{O}}_4}^{3-}$

From the given balanced equation, the mole to mole ratio between Na₃PO₄ and Na⁺ is $1:3$.

So, by using dimensional analysis, we can calculate the moles of Na⁺ present in $1$ mol of Na₃PO₄ as follows:

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}_3{\mathrm{P}\mathrm{O}}_4\mathrm{*}\frac{3\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}_3{\mathrm{P}\mathrm{O}}_4}=3\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}$

So, $1$ mole of Na₃PO₄ contains $3$ mole of Na⁺.

d) $1$ mole of NaNO₃

The dissociation of NaNO₃ from Na⁺ and NO₃^-can be given as below.

${\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}}_3\to \mathrm{ }{\mathrm{N}\mathrm{a}}^{+}+\mathrm{ }{{\mathrm{N}\mathrm{O}}_3}^{-}$

From the given balanced equation, the mole to mole ratio between NaNO₃ and Na⁺ is $1:1$.

So, by using dimensional analysis, we can calculate the moles of Na⁺ present in 1 mol of NaNO₃ as follows:

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}}_3\mathrm{*}\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}}_3}=1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{a}}^{+}$

So, $1$ mole of NaNO₃ contains $1$ mole of Na⁺.

Conclusion:

We have calculated the number of moles of Na⁺ ions from the mole of given compounds and mole ratio.