Chemistry - Mole Concept Question with Solution | TestHub

(P) ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$

1 mole of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ constitutes 8 moles of oxygen atoms.

So in 0.5 mole of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\to 4$ mole of oxygen atoms

(Q) Carbon atom in 112 lit ${\mathrm{CH}}_4$ at NTP.

Mole of ${\mathrm{CH}}_4=\frac{112}{22.4}\Rightarrow \frac{112\times 10}{224}=5$ mole.

1 mole ${\mathrm{CH}}_4\to 1$ mole of Carbon atom

5 mole ${\mathrm{CH}}_4\to 5$ mole of Carbon atoms

(R) ${\mathrm{n}}_{{\mathrm{SO}}_2}=3$; number of moles of Sulphur $=3$