Đáp án+ Giải thích các bước giải:

$a)$

$\text{$m_{KOH}$(20%)}$= $\frac{300.20}{100}$= $60(g)$

$\text{$m_{KOH}$(15%)}$= $\frac{100.15}{100}$= $15(g)$

$m_{dd(sau)}$= $300+100= 400(g)$

$m_{KOH(sau)}$= $60+15=75(g)$

$\rightarrow$ $\text{ $C_{dd (mới)}$ = $\frac{75.100}{400}$= 18,75% }$

$b)$

$n_{H_2SO_4}$= $0,2.1,5+0,5.0,5= 0,55(mol)$

$V_{dd(sau)}$= $0,2+ 0,5=0,7(l)$

$\rightarrow$ $C_{Mdd(mới)}$= $\frac{0,55}{0,7}$= $0,785M$

$\color{cadetblue}{\large{@Sandy}}$
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$\color{darkgoldenrod}{a,}\ m_{KOH\ (dd\ 1)}=\dfrac{300.20}{100}=60\ (g)$

$m_{KOH\ (dd\ 2)}=\dfrac{100.15}{100}=15\ (g)$

$C_{\%\ dd\ mới}=\dfrac{60+15}{300+100}=18,75\%$
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$\color{darkgoldenrod}{b,}\ n_{H_2SO_4\ (dd\ 1)}=1,5.0,2=0,3\ (mol)$

$n_{H_2SO_4\ (dd\ 2)}=0,5.0,5=0,25\ (mol)$

$C_{M\ dd\ mới}=\dfrac{0,3+0,25}{0,2+0,5}=\dfrac{11}{14}M$