Question

What is the molar mass of diacidic organic base, if 12 g of chloroplatinate salt on ignition produced 5 g residue?

Answer 1

When the organic base is reacted with chloroplatinic acid it forms chloroplatinate salt. Salt on further heating forms elemental platinum metal. The reactions of ignition of chloroplatinate are as shown below,
$\begin{aligned} & \text{ B(organic base) + }{{\text{H}}_{\text{2}}}\text{PtC}{{\text{l}}_{\text{6}}}\text{ }\to \text{ B}{{\text{H}}_{\text{2}}}\text{PtC}{{\text{l}}_{\text{6 }}} \\\ & \text{B}{{\text{H}}_{\text{2}}}\text{PtC}{{\text{l}}_{\text{6 }}}\xrightarrow{\text{ignition}}\text{ Pt} \\\ \end{aligned}$

Complete Solution :
We are provided the following data in question:
The weight of chloroplatinate is 12 g.
On ignition produced 5 g of residue.
- We are interested in determining the molar mass of the diacidic organic Lewis base.
Chloroplatinate salt is a platinum complex. it has a complex formula $\,\text{(B}{{\text{H}}_{\text{2}}}\text{)}\left[ \text{PtC}{{\text{l}}_{\text{6}}} \right]\text{ }$.
Here 12 g of chloroplatinate salt undergoes the ignition reaction and produces 5g of platinum residue.
Then 195 g of platinum will correspond to the,
$\text{ weight of chloroplatinate = }\dfrac{12}{5}\times 195\text{ = 468 g }$

Thus 195 g of the platinum will correspond to the 468 g of chloroplatinate complex. Thus 195 $\text{ g/ mol }$ platinum element is present in 468 $\text{ g/ mol }$ chloroplatinate salt. We want to determine that molar mass of the base i.e. $\text{ B}{{\text{H}}_{\text{2 }}}$ .The molar mass of $\text{ B}{{\text{H}}_{\text{2 }}}$ the group would be equal to the difference in the molar mass of chloroplatinate salt and molar mass of $\left[ \text{PtC}{{\text{l}}_{\text{6}}} \right]\text{ }$. The molar mass of $\left[ \text{PtC}{{\text{l}}_{\text{6}}} \right]\text{ }$ is $\text{ 409}\text{.81 g/ mol }$.Thus molar mass of diacidic Lewis base i.e. $\text{ B}{{\text{H}}_{\text{2 }}}$ is as follows,
$\begin{aligned} & \text{ molar mass of B}{{\text{H}}_{\text{2 }}}=\text{ molar mass of chloroplatinate salt}-\text{ molar mass of }\left[ \text{PtC}{{\text{l}}_{\text{6}}} \right]\text{ } \\\ & \Rightarrow \text{molar mass of B}{{\text{H}}_{\text{2 }}} = 468 - 409.81 = 58\text{ g/mol } \\\ \end{aligned}$
Thus molar mass of diacidic organic Lewis base i.e. $\text{ B}{{\text{H}}_{\text{2 }}}$ group is $58\text{ g/mol }$.
So, the correct answer is “Option B”.

Note: Note that molecular mass can be determined by the chloroplatinate method. Here the known amount of organic base is allowed to react with chloroplatinic acid to form chloroplatinate salt. The molar mass can be determined through a direct formula which is given as follows, $\text{ molar mass of base = }\dfrac{1}{2}\left[ \dfrac{\text{weight of chloroplatinate}}{\text{weight of platinum}}\times 195-410 \right]\times \text{Acidity of base }$
- Here the given organic base is diacidic. Thus acidity of the base is 2.