How many litres of ${\mathrm{Cl}}_2$ at STP will be liberated by the oxidation of NaCl with 10 g ${\mathrm{KMnO}}_4$ in acidic medium : (Atomic weight : Mn = 55 and K = 39)

$\begin{array}{l}{\mathrm{Cl}}^{-}+{\mathrm{MnO}}_4^{-}\stackrel{ }{\to }{\mathrm{Mn}}^{+2}+{\mathrm{Cl}}_2\\ \mathrm{Change} \mathrm{in} \mathrm{oxidation} \mathrm{state} \mathrm{of} \text{'}\mathrm{Cl}\text{'} =2\\ \mathrm{Change} \mathrm{in} \mathrm{oxidation} \mathrm{state} \mathrm{of} \text{'}\mathrm{Mn}\text{'} =5\\ \mathrm{Eq}\mathrm{of}{\mathrm{Cl}}_2=\mathrm{Eq}\mathrm{of}{\mathrm{KMnO}}_4\\ 2\left[\mathrm{mole}\mathrm{of}{\mathrm{Cl}}_2\right]=5\left[\frac{10}{158}\right]\\ \mathrm{mole}\mathrm{of}{\mathrm{Cl}}_2=\frac{50}{2\times 158}=0.15823\mathrm{mole}\\ \mathrm{Volume}\mathrm{of}{\mathrm{Cl}}_2\mathrm{at}\mathrm{STP}=0.15823\times 22.4=3.54\mathrm{L}\end{array}$