a,
$C\%= \dfrac{4.100}{4+80}= 4,8\%$
b,
Giả sử có 100ml = 0,1l dd NaOH 2M
$n_{NaOH}= 0,1.2=0,2 mol$
=> $m_{NaOH}= 0,2.40= 8g$
$m_{dd NaOH}= 100.1,08= 108g$
=> $C\%_{NaOH}= \frac{8.100}{108}= 7,4\%$
c,
$m_{dd NaOH}= 3000.1,115= 3345g$
=> $m_{NaOH}= 3345.10\%= 334,5g$
