Mole Concept

The number of oxygen atoms in $4.4\mathrm{\hspace{0.17em}}\mathrm{g}$ of ${\mathrm{CO}}_2$ is:

The number of oxygen atoms in $4.4$ gram of $\text{C}{\text{O}}_2$ is
(Given that atomic mass $\text{C}$ and $\text{O}$ are $\text{12}$ and $\text{16}$ $\text{g/mol}$ )

The number of atoms in $0.1$ moles of a triatomic gas is:

 $({\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}\text{}{\mathrm{mol}}^{-1})$

$1\hspace{0.17em}\mathrm{cc} {\mathrm{N}}_2\mathrm{O}\mathrm{\hspace{0.17em}}\mathrm{at}\mathrm{\hspace{0.17em}}\mathrm{NTP}$ contains:

Which has maximum number of molecules?

If $N_A$ is Avogadro’s number then number of valence electrons in 4.2 g of nitride ions  $(N^{\mathrm{}3-})is$

The volume occupied by one molecule of water$(\mathrm{density}=1 \mathrm{g} {\mathrm{cm}}^{-3})$ is $3\times {10}^{-23} {\mathrm{cm}}^3$.

$600 \mathrm{mL}$ of a mixture of ${\mathrm{O}}_3$ and ${\mathrm{O}}_2$ weighs $1 \mathrm{g}$ at $\mathrm{NTP}$. Calculate the volume of ozone (in mL) in the mixture. (Report your answer in the nearest multiple of 100)

Find the total number of nucleons present in $12 \mathrm{g}$ of ${}^{12}\mathrm{C}$ atoms.

Divide the answer by $1\times {10}^{23}$.

(${\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}$)

Give your answer in two significant figures.

Calculate the mass of $5$ moles of ${\mathrm{CaCO}}_3$ in grams.

How many moles of electrons weigh one kilogram?

A mixture contains $5.4 \mathrm{g}$ of $\mathrm{Al}$, $1.2 \mathrm{g}$ of $\mathrm{Mg}$ and $4.6 \mathrm{g}$ of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$. The ratio of their moles is

(Atomic weights of $\mathrm{Al}=27 \mathrm{u}, \mathrm{Mg}=24 \mathrm{u}, \mathrm{C}=12 \mathrm{u}, \mathrm{O}=16 \mathrm{u}, \mathrm{H}=1 \mathrm{u}$)

The volume occupied by one molecule of water $(\mathrm{density}=1 \mathrm{g} {\mathrm{cm}}^{-3})$ is:

The volume occupied by one molecule of water$(\mathrm{density}=1 \mathrm{g} {\mathrm{cm}}^{-3})$ is $3\times {10}^{-23} {\mathrm{cm}}^3$.