(4)

A.   No. of atoms in 212 g of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ $=\frac{6\times N_A\times 212}{106}=12N_A$

B.   No. of atoms in 248 g of ${\mathrm{Na}}_2\mathrm{O}=\frac{3\times {\mathrm{N}}_{\mathrm{A}}\times 248}{62}=12{\mathrm{N}}_{\mathrm{A}}$

C.   No. of atoms in 240 g of NaOH $=\frac{3\times N_A\times 240}{40}=18N_A$

D.   No. of atoms in 12 g of ${\mathrm{H}}_2=\frac{2\times {\mathrm{N}}_{\mathrm{A}}\times 12}{2}=12{\mathrm{N}}_{\mathrm{A}}$

E.   No. of atoms in 220 g of ${\mathrm{CO}}_2=\frac{3\times N_A\times 220}{44}=15N_A$

$\therefore$   A, B and D have equal number of atoms.

(1)

(1)

4 mol of helium $=4\times N_A$ atoms

4 u of helium = 1 atom

4 g of helium = 1 mol = $N_A$ atoms

2.271098 L of helium $=\frac{2.271098}{22.7}$mol = 0.1 mol = 0.1 $N_A$ atoms

(4)

1 mole of substance $=N_A$ atoms

108 g of Ag $=N_A$ atoms $\Rightarrow$ 1 g of Ag $=\frac{N_A}{108}$ atoms

24 g of Mg $=N_A$ atoms $\Rightarrow$ 1 g of Mg $=\frac{N_A}{24}$ atoms

32 g of ${\text{O}}_2=N_A$ molecules $=2N_A$ atoms

$\Rightarrow$ 1 g of $O_2=\frac{N_A}{16}$ atoms

7 g of Li $=N_A$ atoms $\Rightarrow$ 1 g of Li $=\frac{N_A}{7}$ atoms

Therefore, 1 g of ${\mathrm{Li}}_{\left(\mathrm{s}\right)},$ has maximum number of atoms.

(1)

(1)  Mass of water $=V\times d=18\times 1=18$$g$

      Molecules of water $=\text{mole}\times N_A=\frac{18}{18}{N}_A=N_A$

(2)  Molecules of water $=\text{mole}\times N_A=\frac{0.18}{18}{N}_A={10}^{-2}{N}_A$

(3)  Moles of water $=\frac{0.00224}{22.4}={10}^{-4}$

       Molecules of water $=\text{mole}\times N_A={10}^{-4}{N}_A$

(4)  Molecules of water $=\text{mole}\times N_A={10}^{-3}{N}_A$

(1)

Let atomic weight of element X be x and that of element Y be y.

For $X{Y}_{\mathit{2}}\mathit{,}$ $n=\frac{w}{\text{Mol. wt.}}$

$0.1=\frac{10}{x+2y}\Rightarrow x+2y=\frac{10}{0.1}=100$                             ...(i)

For ${\mathrm{X}}_3{\mathrm{Y}}_2,$ $n=\frac{w}{\text{Mol. wt.}}$

$0.05=\frac{9}{3x+2y}\Rightarrow 3x+2y=\frac{9}{0.05}=180$                   ...(ii)

On solving equations (i) and (ii), we get $x=40$

$40+2y=100\Rightarrow 2y=60\Rightarrow y=30$

(1)

Mass of 1 mol $(6.022\times {10}^{23}\text{ atoms})$ of carbon = 12 g

If Avogadro number is changed to $6.022\times {10}^{20}$ atoms, then mass of 1 mol of carbon

         $=\frac{12\times 6.022\times {10}^{20}}{6.022\times {10}^{23}}=12\times {10}^{-3}$$\mathrm{g}$

(3)

1.8 gram of water $=\frac{6.023\times {10}^{23}}{18}\times 1.8$

                              $=6.023\times {10}^{22}\text{ molecules}$

18 gram of water $=6.023\times {10}^{23}$ molecules

18 moles of water $=18\times 6.023\times {10}^{23}$ molecules

(4)

Number of moles of ${\mathrm{H}}_2=\frac{1}{2}$

Number of moles of ${\mathrm{O}}_2=\frac{4}{32}$

Hence, molar ratio $=\frac{1}{2}:\frac{4}{32}=4:1$