a) Đặt \(V_{d^2\text{ }A}=x\left(l\right)\)
\(\Rightarrow V_{d^2\text{ }B}=1,5x\left(l\right)\\ \Rightarrow V_{d^2\text{ }C}=1,5x+x=2,5x\left(l\right)\)
\(\Rightarrow n_{H_2SO_4\text{ }trong\text{ }A}=V\cdot C_M=0,2x\left(mol\right)\\ n_{H_2SO_4\text{ }trong\text{ }B}=V\cdot C_M=1,5x\cdot0,5=0,75x\left(mol\right)\\ \Rightarrow n_{H_2SO_4\text{ }trong\text{ }C}=0,2x+0,75x=0,95x\left(mol\right)\)
\(\Rightarrow C_{M\left(C\right)}=\dfrac{n}{V}=\dfrac{0,95x}{2,5x}=0,38\left(M\right)\)
b) Đặt \(V_A=a\left(l\right)\)
\(V_B=b\left(l\right)\\ \Rightarrow V_{d^2\text{ }cần\text{ }pha\text{ }chế}=a+b\left(l\right)\\ \Rightarrow n_{d^2\text{ }cần\text{ }pha\text{ }chế}=C_M\cdot V=0,3\left(a+b\right)\left(mol\right)\)
\(\Rightarrow n_A=C_M\cdot V=0,2a\left(mol\right)\\ n_B=C_M\cdot V=0,5b\left(mol\right)\)
\(\text{Ta có }pt:0,2a+0,5b=0,3\left(a+b\right)\\ \Leftrightarrow0,2a+0,5b=0,3a+0,3b\\ \Leftrightarrow0,2b=0,1a\\ \Rightarrow\dfrac{a}{b}=\dfrac{0,2}{0,1}=2:1\)
Vậy phải trộn theo tỉ lệ \(V_A:V_B=2:1\) để được dung dịch \(0,3M\)