1.

`d_{SO_2//Cl_2}={M_{SO_2}}/{M_{Cl_2}}={64}/{71}\approx 0,901`

`->` Tỉ khối `SO_2` đối với `Cl_2` khoảng `0,901` 

2.

`M_{H_2O}=2M_H+M_O=2.1+16=18(a m u)`

3.

`m_{dd\ NaCl}=m_{NaCl}+m_{H_2O}=20+50=70(g)`

`->C\%_{NaCl}={m_{NaCl}}/{m_{dd\ NaCl}}.100\%={20}/{70}.100\%\approx 28,57\%`

4.

`S_{NaNO_3}={m_{NaNO_3}}/{m_{H_2O}}.100={12}/{15}.100=80(g)`

`1,`

`d_(\SO_2//_(Cl_2}` `=M_{SO_2}/M_(Cl_2}=` `64/71~~0,90\ l ầ n.`

`->\ SO_2` đối với `Cl_2` khoảng bằng `0,90.` lần`.`

`2,`

`M_{H_2O}=2xx1+16=18(g//mol)`

`3,`

`m_{dd\ NaCl}=m_{NaCl}+m_{H_2O}=20+50=70(g//mol)`

`-> C_{M\ NaCl}=m_{ct}/m_{dd}xx100=20/70 xx 100~~28,6%`

`4)`

`S_{NaNO_3}=m_{ct}/m_{dm} xx100=12/15 xx 100=80(g)`