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(D) To find the number of atoms,we calculate the moles of molecules and multiply by the number of atoms per molecule ($N_A$ is Avogadro's number).$A)

Vedclass · Accepted Answer

$1 \ mL$ of $H_2O$

Vedclass · Answer

(D) To find the number of atoms,we calculate the moles of molecules and multiply by the number of atoms per molecule ($N_A$ is Avogadro's number).$A) \ 1 \ mg \ C_4H_{10} = \frac{10^{-3} \ g}{58 \ g/mol} 	imes 14 \ atoms 	imes N_A \approx 0.24 	imes 10^{-3} \ N_A \ atoms$$B) \ 1 \ mg \ N_2 = \frac{10^{-3} \ g}{28 \ g/mol} 	imes 2 \ atoms 	imes N_A \approx 0.07 	imes 10^{-3} \ N_A \ atoms$$C) \ 1 \ mg \ Na = \frac{10^{-3} \ g}{23 \ g/mol} 	imes 1 \ atom 	imes N_A \approx 0.04 	imes 10^{-3} \ N_A \ atoms$$D) \ 1 \ mL \ H_2O = 1 \ g \ H_2O = \frac{1 \ g}{18 \ g/mol} 	imes 3 \ atoms 	imes N_A \approx 0.166 \ N_A \ atoms$Comparing the values,$1 \ mL$ of $H_2O$ contains the largest number of atoms.

Column-$I$	Column-$II$
$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$	$I$. $32 \ g$
$B$. $1.2 \times 10^{24}$ atoms of oxygen	$II$. $66 \ g$
$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$	$III$. $42 \ g$
$D$. $1.5 \ g$-molecule of $N_2$	$IV$. $16 \ g$

Which sample contains the largest number of atoms?

Similar Questions

An organic compound contains $46.78 \%$ of a halogen $X$. When $2.00 \ g$ of this compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$,$2.21 \ g$ of $AgX$ is formed. The halogen $X$ is:
[Given: atomic weight of $Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127$]

Identify the type of reaction for the following chemical equation:
$CuSO_4 + 2NaOH (excess) \longrightarrow Cu(OH)_2 \downarrow + Na_2SO_4$

Match the following:

Column-$I$ Column-$II$

$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$ $I$. $32 \ g$

$B$. $1.2 \times 10^{24}$ atoms of oxygen $II$. $66 \ g$

$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ $III$. $42 \ g$

$D$. $1.5 \ g$-molecule of $N_2$ $IV$. $16 \ g$

The correct match is:

The ratio of $C_p$ and $C_v$ of a gas '$X$' is $1.4$. The number of atoms of the gas '$X$' present in $11.2 \ L$ of it at $N.T.P.$ is

Lattice enthalpy and enthalpy of solution of $NaCl$ are $788 \ kJ \ mol^{-1}$ and $4 \ kJ \ mol^{-1}$ respectively. The hydration enthalpy of $NaCl$ is ....... $kJ \ mol^{-1}$

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