Fill in the missing data in the table

		2.	3.	4.
Species property	${\mathrm{H}}_2\mathrm{O}$	${\mathrm{CO}}_2$	$\mathrm{Na}$ atom	${\mathrm{MgCl}}_2$
No of Moles	$2$	(i)______	(ii)______	$0.5$
No of particles	(iii)_____	$3.011\times {10}^{23}$	(iv)_______	(v)_____
Mass	$36\mathrm{g}$	(vi) _______	$115\mathrm{g}$	(vii)______

1. ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Part-$1$: Given data:

• Number of moles$\left(\mathrm{n}\right)$ $=$ $2$
• mass of the compound$\left(\mathrm{m}\right)=36\mathrm{g}$

Part-$2$: Calculating the number of particles:

• Number of particles $=$$\mathrm{n}\times {\mathrm{N}}_{\mathrm{A}}$ ; $({\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23})$
• Number of particles $$\begin{gathered} =2\times 6.022\times {10}^{23} \\ =12.044\times {10}^{23} \end{gathered}$$

2. ${\mathbf{CO}}_{\mathbf{2}}$

Part-$1$: Given data:

• Number of particles $=3.011\times {10}^{23}$

Part-$2$: Calculating the number of moles$\left(\mathrm{n}\right)$

• Number of particles $=\mathrm{number}\mathrm{of}\mathrm{moles}\times {\mathrm{N}}_{\mathrm{A}}$
• $$\begin{gathered} \Rightarrow 3.011\times {10}^{23}=n\times 6.022\times {10}^{23} \\ \Rightarrow n=0.5mol \end{gathered}$$

Part-$3$: Calculating the mass of the compound$\left(\mathrm{m}\right)$

• Number of moles$\left(\mathrm{n}\right)$ $=\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{compound}\left(\mathrm{m}\right)}{Molarmassofthecompound\left(M\right)}$
• Molar mass of ${\mathrm{CO}}_2$ $(\mathrm{C}=12;\mathrm{O}=16)$$\Rightarrow 12+(16\times 2)=44gmo{l}^{-1}$
• $$\begin{gathered} \Rightarrow 0.5=\frac{m}{44} \\ \Rightarrow m=44\times 0.5 \\ \Rightarrow m=22g \end{gathered}$$

3. $\mathrm{Na}$ atom

Part-$1$: Given data:

• mass of the element$\left(\mathrm{m}\right)=115\mathrm{g}$
• Molar mass of $\mathrm{Na}=23{\mathrm{gmol}}^{-1}$

Part-$2$: Calculating the number of moles$\mathbf{\left(}\mathbf{n}\mathbf{\right)}$

• Number of moles$\left(\mathrm{n}\right)$ $=\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{compound}\left(\mathrm{m}\right)}{Molarmassofthecompound\left(M\right)}$
• $$\begin{gathered} \Rightarrow n=\frac{115}{23} \\ \Rightarrow n=5mol \end{gathered}$$

Part-$3$: Calculating the number of particles:

• Number of particles $=\mathrm{n}\times {\mathrm{N}}_{\mathrm{A}}$
• $$\begin{gathered} Numberofparticles=5\times 6.022\times {10}^{23} \\ =30.11\times {10}^{23} \\ =3.011\times {10}^{24}particles. \end{gathered}$$

4. ${\mathbf{MgCl}}_{\mathbf{2}}$

Part-$1$: Given data:

• Number of moles$\left(\mathrm{n}\right)=0.5$
• Molar mass of ${\mathrm{MgCl}}_2$$$\begin{gathered} =24.3+(35.5\times 2) \\ =95.3{\mathrm{gmol}}^{-1} \end{gathered}$$

Part-$2$: Calculating the number of particles:

• Number of particles $=\mathrm{n}\times {\mathrm{N}}_{\mathrm{A}}$
• $$\begin{gathered} Numberofparticles=0.5\times 6.022\times {10}^{23} \\ =3.011\times {10}^{23}particles. \end{gathered}$$

Part-$3$: Calculating the mass of compound $\mathbf{\left(}\mathbf{m}\mathbf{\right)}$

• Number of moles$\left(\mathrm{n}\right)$ $=\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{compound}\left(\mathrm{m}\right)}{Molarmassofthecompound\left(M\right)}$
• $$\begin{gathered} \Rightarrow 0.5=\frac{m}{95.3} \\ \Rightarrow m=47.65g \end{gathered}$$

So,

		2.	3.	4.
Species property	${\mathrm{H}}_2\mathrm{O}$	${\mathrm{CO}}_2$	$\mathrm{Na}$ atom	${\mathrm{MgCl}}_2$
No of Moles	$2$	$\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{mol}$	$\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\mathbf{5}\mathbf{}\mathbf{mol}$	$0.5$
No of particles	$\mathbf{\left(}\mathbf{iii}\mathbf{\right)}\mathbf{}\mathbf{12}\mathbf{.}\mathbf{044}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{particles}$	$3.011\times {10}^{23}$	$\mathbf{\left(}\mathbf{iv}\mathbf{\right)}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{011}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathbf{particles}$	$\mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{011}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{particles}$
Mass	$36\mathrm{g}$	$\mathbf{\left(}\mathbf{vi}\mathbf{\right)}\mathbf{}\mathbf{22}\mathbf{g}$	$115\mathrm{g}$	$\mathbf{\left(}\mathbf{vii}\mathbf{\right)}\mathbf{}\mathbf{47}\mathbf{.}\mathbf{65}\mathbf{g}$