$1)n_{CO_2}=\dfrac{1,83.10^{22}}{6.10^{23}}=0,0305(mol)\\ \Rightarrow V_{CO_2(20^oC,0,25atm)}=\dfrac{n.R.T}{P}=\dfrac{0,0305.0,082.(273+20)}{0,25}=0,18319825(l)\\ 2)840(ml)=0,84(l)\\ \Rightarrow n_{N_2}=\dfrac{P.V}{R.T}=\dfrac{4.0,84}{0,082.(273+20)}\approx 0,14(mol)\\ \Rightarrow m_{N_2}=n.M=0,14.28=3,92(g)\\ 3)112(ml)=0,112(l)\\ \Rightarrow n_{Cl_2}=\dfrac{P.V}{R.T}=\dfrac{2.0,112}{0,082.273}\approx 0,01(mol)\\ \Rightarrow m_{Cl_2}=0,01.71=0,71(g)\\ 4)n_{O_2}=\dfrac{1,26.10^{21}}{6.10^{23}}=0,0021(mol)\\ \Rightarrow m_{O_2}=0,0021.32=0,0672(g)\\ 5)\text{Đặt }n_{N_2}=a(mol)\\ \Rightarrow m_{N_2}=28a(g)\\ \Rightarrow m_{CO_2}=\dfrac{33}{7}m_{N_2}=132a(g)\\ \Rightarrow n_{CO_2}=\dfrac{132a}{44}=3a(mol)\\ \Rightarrow M_B=\dfrac{m_{CO_2}+m_{N_2}}{n_{CO_2}+n_{N_2}}=\dfrac{132a+28a}{3a+a}=40(g/mol)\\ \Rightarrow d_{B/O_2}=\dfrac{40}{32}=1,25$