91Ó°ÊÓ

We have to calculate the molar mass of ${\mathrm{O}}_2$

We know that,

Molar mass of one $\mathrm{O}$atom =localid="1653485525043" $15.99\mathrm{g}$

So, molar mass of ${\mathrm{O}}_2$( two $\mathrm{O}$atoms ) =$2\left(15.99\right)$=localid="1653485535668" $31.98\mathrm{g}$

We have to calculate the molar mass of ${\mathrm{KH}}_2{\mathrm{PO}}_4$

We know that,

Molar mass of one $\mathrm{K}$atom =localid="1653485549986" $39.09\mathrm{g}$

Molar mass of one $\mathrm{H}$atom =localid="1653485559971" $1.00\mathrm{g}$

Molar mass of one $\mathrm{P}$atom =localid="1653485572328" $30.97\mathrm{g}$

Molar mass of one$\mathrm{O}$atom =localid="1653485593656" $15.99\mathrm{g}$

So, molar mass of ${\mathrm{KH}}_2{\mathrm{PO}}_4$=$39.09+2\left(1.00\right)+30.97+4\left(15.99\right)$=localid="1653485613125" $136.02\mathrm{g}$

We have to calculate the molar mass of $\mathrm{Fe}{\left({\mathrm{ClO}}_4\right)}_3$

We know that,

Molar mass of one $\mathrm{Fe}$atom =localid="1653485626966" $55.84\mathrm{g}$

Molar mass of one$\mathrm{O}$atom =localid="1653485634822" $15.99\mathrm{g}$

Molar mass of one $\mathrm{Cl}$atom =localid="1653485644819" $35.45\mathrm{g}$

So, molar mass of $\mathrm{Fe}{\left({\mathrm{ClO}}_4\right)}_3$=$55.84+3\left(35.45\right)+12\left(15.99\right)$=localid="1653485655673" $354.07\mathrm{g}$