Question

i) Which has more number of atoms? (5M)

a) 10 gram of nitrogen (N2).

b) 10 gram of ammonia (NH3)

ii) Calculate the total number of moles in 0.585 gram of sodium chloride. (Atomic mass

of N=14u ,H=1u, Cl=35.5u & Na=23u.).​

Answer 1

1) 28g of Dinitrogen gas has 6.022 × 10²³ Atoms.

In other words, 28g of $\sf N_2$ has 1 mol of atoms.

Let 10g of $\sf N_2$ has x number of atoms.

$\longrightarrow \sf \: x = \dfrac{10}{28} \\ \\ \longrightarrow \sf \: x \sim \: 0.35 \: moles$

Thus, 10g of $\sf N_2$ has 0.35 mol.

17g of $\sf NH_3$ has 1 mole.

Let 10g of $\sf NH_3$ have y number of moles.

$\longrightarrow \sf \: y= \dfrac{10}{17} \\ \\ \longrightarrow \sf \: y \sim \: 0.58\: moles$

Thus, 10g of $\sf NH_3$ has 0.58 mol.

10g of $\sf NH_3$ has more number of atoms.

2) Mass of NaCl = 1 × 23 + 1 × 35.5 = 58.5 grams.

Total number of atoms is given as :

$\sf n = \dfrac{Given \ Mass}{Molar \ Mass}$

Thus,

$\longrightarrow \sf \: n= \dfrac{0.585}{58.5} \\ \\ \longrightarrow \sf \: n = \dfrac{585 \times {10}^{ - 3} }{585 \times {10}^{ - 1} } \\ \\ \longrightarrow \boxed{ \boxed{\sf \: n = {10}^{ - 2} \: moles}}$

Answer 2

1) 28g of Dinitrogen gas has 6.022 × 10²³ Atoms.

In other words, 28g of \sf N_2N

2

has 1 mol of atoms.

Let 10g of \sf N_2N

2

has x number of atoms.

\begin{gathered} \longrightarrow \sf \: x = \dfrac{10}{28} \\ \\ \longrightarrow \sf \: x \sim \: 0.35 \: moles\end{gathered}

⟶x=

28

10

⟶x∼0.35moles

Thus, 10g of \sf N_2N

2

has 0.35 mol.

17g of \sf NH_3NH

3

has 1 mole.

Let 10g of \sf NH_3NH

3

have y number of moles.

\begin{gathered} \longrightarrow \sf \: y= \dfrac{10}{17} \\ \\ \longrightarrow \sf \: y \sim \: 0.58\: moles\end{gathered}

⟶y=

17

10

⟶y∼0.58moles

Thus, 10g of \sf NH_3NH

3

has 0.58 mol.

10g of \sf NH_3NH

3

has more number of atoms.

2) Mass of NaCl = 1 × 23 + 1 × 35.5 = 58.5 grams.

Total number of atoms is given as :

\sf n = \dfrac{Given \ Mass}{Molar \ Mass}n=

Molar Mass

Given Mass

Thus,

\begin{gathered} \longrightarrow \sf \: n= \dfrac{0.585}{58.5} \\ \\ \longrightarrow \sf \: n = \dfrac{585 \times {10}^{ - 3} }{585 \times {10}^{ - 1} } \\ \\ \longrightarrow \boxed{ \boxed{\sf \: n = {10}^{ - 2} \: moles}}\end{gathered}

⟶n=

58.5

0.585

⟶n=

585×10

−1

585×10

−3

⟶

n=10

−2

moles