1.

`a)`

`m_{NaOH}=50.10\%=5(g)`

`->C\%_{NaOH\ sau}=5/{50+20}.100\%\approx 7,14\%`

`b)` `C\%_{NaOH\ sau}=5/{40}.100\%=12,5\%`

2.

`a)`

`100mL=0,1L;20mL=0,02L`

`n_{H_2SO_4}=0,1.1=0,1(mol)`

`->C_{M\ H_2SO_4\ sau}={0,1}/{0,1+0,02}\approx 0,83M`

`b)`

Sửa: cô đặc còn lại `40mL\ H_2SO_4`

`40mL=0,04L`

`->C_{M\ H_2SO_4\ sau}={0,1}/{0,04}=2,5M`

4.

`m_{NaOH\ 20\%}=60.20\%=12(g)`

`m_{NaOH\ 15\%}=40.15\%=6(g)`

`->C\%_{NaOH\ sau}={12+6}/{60+40}.100\%=18\%`