`@` $KOH$ cần pha.

$m_{dd,KOH}=1.5 \times 10^3 \times 1.1 =1650(gam)$

$m_{KOH}=8\%m_{dd,KOH}=8\% \times 1650=132(gam)$

$n_{KOH}=\dfrac{132}{56}=\dfrac{33}{14}(mol)$

`@` $KOH$ $4\%$ 

$m_{dd,KOH}=V_1 \times 1.05=1.05V_1 (gam)$

$m_{KOH}=4\%m_{dd,KOH}=4\% \times V_1 \times 1.05=0.042V_1(gam)$

$n_{KOH}=\dfrac{0.042V_1}{56}=0.00075V_1(mol)$

`@` $KOH$ $4\%$

Đặt $V_2=(1.5 \times 10^3-V_1)$ 

$m_{dd,KOH}=V_2 \times 1.12 =1680-1.12V_1 (gam)$

$m_{KOH}=10\%m_{dd,KOH}=10\% \times (1680-1.12V_1) =168-\dfrac{1.12V_1}{10} (gam)$

$n_{KOH}=\dfrac{168-\dfrac{1.12V_1}{10} }{56}=\dfrac{1680-1.12V_1}{560}(mol)$

Bảo toàn số mol $KOH:$

`->`$0.00075V_1+\dfrac{1680-1.12V_1}{560}=\dfrac{33}{14}$

`=>`$V_1=\dfrac{360}{0.7}≈514(mL)$

`=>`$V_2=1.5 \times 10^3-514=986(mL)$

Đáp án:

Giải thích các bước giải: