91Ó°ÊÓ

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

We know that,

Molar mass of one $\mathrm{Cl}$atom = localid="1653485335070" $35.45\mathrm{g}$

So, molar mass of ${\mathrm{Cl}}_2$( two $\mathrm{Cl}$atoms ) = $2×\left(35.45\right)$=localid="1653485344916" $70.90\mathrm{g}$

We have to calculate the molar mass of ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$.

We know that,

The molar mass of one $\mathrm{C}$atom = localid="1653485362731" $12.01\mathrm{g}$

The molar mass of one $\mathrm{H}$atom =localid="1653485373603" $1.00\mathrm{g}$

The molar mass of one $\mathrm{O}$atom = localid="1653485380629" $15.99\mathrm{g}$

So, the molar mass of ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$= $3\left(12.01\right)+6\left(1.00\right)+3\left(15.99\right)$= localid="1653485392293" $90\mathrm{g}$

We have to calculate the molar mass of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$.

We know that,

The molar mass of one $\mathrm{Mg}$atom = localid="1653485409171" $24.30\mathrm{g}$

The molar mass of one $\mathrm{P}$atom =localid="1653485419047" $30.97\mathrm{g}$

The molar mass of one $\mathrm{O}$atom =localid="1653485426714" $15.99\mathrm{g}$

So, the molar mass of${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$=$3\left(24.30\right)+2\left(30.97\right)+8\left(15.99\right)$=localid="1653485434208" $262.76\mathrm{g}$