`a)` `S=\frac{m_{KNO_3}}{m_{dm}}.100`
`=> m_{KNO_3}=S.100.\frac{1}{100}`
`=> m_{KNO_3}=31,6g`
`m_{dd}=m_{KNO_3}+m_{dm}=31,6+100=131,6g`
`=> C%_{KNO_3}=\frac{31,6}{131,6}.100%\approx 24%`
`b)` `m_{dd}=180+a(g)`
Ta có:
`C%_{NaOH}=\frac{m_{NaOH}}{m_{dd}}.100%`
`=>20%=\frac{a}{180+a}.100%`
`=> 0,2=\frac{a}{180+a}`
`=>36+0,2a=a`
`=>a=45`