Question

The number of ions present in $1ml$ of $0.1M$ $CaC{l}_2$ solution is:

• A. $1.8\times {10}^{20}$
• B. $6.0\times {10}^{20}$
• C. $1.8\times {10}^{19}$
• D. $1.8\times {10}^{21}$

Answer 1

Answer: (A)

$1.8\times {10}^{20}$

$Molarity=\frac{molesofsolute}{Volumeofsolution\left(mL\right)}\times 1000$

$\therefore molesofsolute=\frac{Molarity\times Volumeofsolution\left(mL\right)}{1000}$

$\therefore MolesofCaC{l}_2=\frac{0.1\times 1}{1000}={10}^{-4}$ $moles$

No. of molecules of $CaC{l}_2=$ Moles $\times$ Avogadro's no.

$={10}^{-4}\times 6.022\times {10}^{23}=6.022\times {10}^{19}$ molecules

$CaC{l}_2⟶C{a}^{2+}+2C{l}^{-}$

$1$ molecule of $CaC{l}_2$ gives $3$ ions.

$\therefore$ No. of ions $=3\times$ no. of molecules of $CaC{l}_2=3\times 6.022\times {10}^{19}$

$=1.8\times {10}^{20}$ $ions$

Hence, option $A$ is correct.