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  $(N^{\mathrm{}3-})is$

1 gm of charcoal adsorbs 100 mL 0.5 M ${\mathrm{CH}}_3\mathrm{COOH}$ to form a monolayer, and thereby the molarity of ${\mathrm{CH}}_3\mathrm{COOH}$ reduces to 0.49. The surface area of the charcoal adsorbed by each molecule of acetic acid. $\left(\text{Surface area of charcoal = 3.01}\times {\text{10}}^2{\text{ m}}^2/\text{gm, Molarity}=\frac{\text{Number of moles}}{\text{1000 mL}}\right)$ will be:

The number of oxygen atoms in $6.4\mathrm{g} \mathrm{of} {\mathrm{SO}}_2$is:​

Calculate the amount of lime, $\mathrm{Ca}(\mathrm{OH}{)}_2$, required to remove hardness of $50,000$ litres of well water which contains$1.62 \mathrm{g}$ of calcium bicarbonate per$10 \mathrm{liters}.$
[Given, atomic masses: $\mathrm{Ca}\to 40,\mathrm{H}\to 1,\mathrm{C}\to 12,0\to 16]$

$3.011\times {10}^{22}$ atoms of an element weigh $1.15\mathrm{gm}$. The atomic mass of the element is

$100 \mathrm{ml}$ of $0.2 \mathrm{M}$ acetic acid is completely neutralized using a standard solution of $\mathrm{NaOH}.$ The volume of ethane obtained at $\mathrm{STP}$ after complete electrolysis of the resulting solution is $\_\_\_\_\_\_\_\_\_\_\_\_$

If one atom of an element $\mathrm{X}$ weighs $6.643\times {10}^{-23} \mathrm{g}.$ Then find the number of moles of atoms in $50 \mathrm{kg}$ of element $\mathrm{X}.$