Calculate the active masses of carbon tetrachloride [density= 1.58 g (mL].

g. mol. wt. of ${\mathrm{CCI}}_4$$=12+(4\times 35.5)=12+142.0=154$

$1\mathrm{L}{\mathrm{CCl}}_4\text{ weighs }=1000\mathrm{mL}\times \text{ density in }\mathrm{g}(\mathrm{mL}{)}^{-}$

$=1000\mathrm{mL}\times 1.58\mathrm{g}(\mathrm{mL}{)}^{-}=1580\mathrm{g}$

$\therefore$ no. of mol of ${\mathrm{CCI}}_4$ 

$=\frac{1580}{154}=10.26{\mathrm{molL}}^{-}$

or Active mass of ${\mathrm{CCI}}_4$

$=10.26{\mathrm{molL}}^{-}$