Question

Calculate the value of Avagadro's number if the density of $NaCl$ is $2.165$ g cm${}^{-3}$ and distance between $N{a}^{+}$ and $C{l}^{-}$ ions in $NaCl$ crystal is $281$ pm.

• A. $1.52\times {10}^{23}$
• B. $3.04\times {10}^{23}$
• C. $6.09\times {10}^{23}$
• D. $12.18\times {10}^{23}$

Answer 1

Answer: (C)

$6.09\times {10}^{23}$

Density of NaCl $=2.165$ g/cc

Number of atoms in NaCl $=4$

FCC structure $r_{N{a}^{+}}+R_{C{l}^{-}}=\frac{a}{2}$

$281=0.3535a\Rightarrow a=794.9$ pm $=794.9\times {10}^{-10}$ cm
Molecular weight of NaCl $=23+35.5=58.5$ g/mol

$d=\frac{z\times m}{a^3\times N_A}$

$2.165=\frac{4\times 58.5}{(794.9\times {10}^{-10}{)}^3\times N_A}$

$N_A=\frac{4\times 58.5}{(794.9\times {10}^{-10}{)}^3\times 2.165}$

$N_A=6.09\times {10}^{23}$

Hence, the value of the avogadro's number is $6.09\times {10}^{23}$.

Hence, the correct option is $\text{C}$