Question

The number of molecules of water in $333g$ of $A{l}_2{\left(S{O}_4\right)}_3\cdot 18H_{2}O$ is

• A. $18.0\times 6.023\times 10^{23}$
• B. $9.0\times 6.023\times 10^{23}$
• C. $18.0$
• D. $36.0$

Answer 1

Answer: (B)

Formula of the compound suggests that one molecule of the compound contains $18$ molecules of water. Number of water molecules in $333g$ of compound

=number of moles $\times 6.023\times 10^{23}\times 18$

$=\frac{333}{666}\times 6.023\times 10^{23}\times 18=9\times 6.023\times 10^{23}$