Chapter 10.5, Problem 81SSC

Interpretation Introduction

Interpretation:

The given hydrates need to be arranged in increasing order of percent water per weight.

Concept introduction:

In the given compound, the mass percent of any component can be calculated as follows:

$\%m_A=\frac{m_A}{m_{compound}}$

Here, $m_A$ is mass of the component and $m_{compound}$ is mass of compound.

Answer to Problem 81SSC

${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O<Ba(OH)}}_{\text{2}}{\text{.8H}}_{\text{2}}{\text{O<MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{1}{\text{. MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}\\ \text{2}{\text{. Ba(OH)}}_{\text{2}}{\text{.8H}}_{\text{2}}\text{0}\\ \text{3}{\text{. CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}\end{array}$

Calculations:

The first hydrate is as follows:

${\text{MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$

Calculate the molar mass of the hydrate as follows:

$M=24+96+7(18)=246\text{ g/mol}$

The molar mass of water is 18 g/mol.

The percentage of water in

${\text{MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$ (by weight) can be calculated as follows: $\%m=\frac{7\times 18}{246}\times 100=51.22\%$

The second hydrate is as follows:

${\text{Ba(OH)}}_{\text{2}}{\text{.8H}}_{\text{2}}\text{O}$

Calculate the molar mass of the hydrate as follows:

$\text{molar mass}=137.327+2(17)+8(18)=315.327\text{ g/mol}$

Percentage of water in $Ba{(OH)}_2.8H_{2}O$ (by weight) can be determined as follows: $\%m=\frac{8\times 18}{315.327}\times 100=45.67\%$

The 3^rd hydrate is as follows:

${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$

The molar mass of the compound $=59+2(35.5)+6(18)=238\text{ g/mol}$

Percentage water in ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$ (by weight) $=\frac{6\times 18}{238}\times 100=45.35\%$

Thus,

So, the correct order is as follows:

${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O<Ba(OH)}}_{\text{2}}{\text{.8H}}_{\text{2}}{\text{O<MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$

Conclusion

Thus, ${\text{MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$ has maximum and ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$ has minimum percentage of water.

${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O<Ba(OH)}}_{\text{2}}{\text{.8H}}_{\text{2}}{\text{O<MgSO}}_{\text{4}}{\text{.7H}}_{\text{2}}\text{O}$