Question

Calculate the number of formula units of
$\left[Ag\right(N{H}_3{)}_2]Cl$ in 1 kilomole of $\left[Ag\right(N{H}_3{)}_2]Cl$.

• A. $6.022\times {10}^{23}$
• B. $3.011\times {10}^{23}$
• C. $6.022\times {10}^{26}$
• D. $6.022\times {10}^{20}$

Answer 1

The correct option is C $6.022\times {10}^{26}$

$\text{Number of formula units = number of moles}\times N_{A}formulaunits/mol$
$1kilomole=1\times {10}^{3}molesor1000moles$
$\text{Number of formula units}=1\times {10}^3\times 6.022\times {10}^{23}=6.022\times {10}^{26}$