\(n_{Cl_2}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25mol\\ m_{Cl_2}=0,25.71=17,75g\\ V_{Cl_2}=0,25.24,79=6,1975l\)
\(n_{Cl_2}=\dfrac{N}{A}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\\ m_{Cl_2}=0,25.71=17,75\left(g\right)\\ V_{Cl_2}=0,25.22,4=5,6\left(l\right)\)