a/ 1/10 dd C ứng với 0,1 lit.
HCl + AgNO3 --> AgCl + HNO3
nHCl = nAgCl = \(\dfrac{8,61}{143,5}\) = 0,06 mol
=> CM(C) = \(\dfrac{0,06}{0,1}\) = 0,6 mol / l
b/ Gọi Ca, Cb là nồng độ mol của ddA và ddB.
gt: Ca = 4.Cb (1)
mặt khác:
nHCl(A) = \(\dfrac{1}{3}\)Ca
nHCl(B) = \(\dfrac{2}{3}\)Cb
nHCl(C) = 1.0,6 = 0,6
ta có:
\(\dfrac{1}{3}\)Ca + \(\dfrac{2}{3}\)Cb = 0,6
=> Ca + 2.Cb = 1,8 (2)
Giải hệ (1), (2):
4.Cb + 2.Cb = 1,8
=>Cb = 0,3 M
Ca = 1,2 M
a/ \(n_{AgCl}=\dfrac{8,61}{143,5}=0,06\left(mol\right)\)
PTHH: HCl + AgNO3 ----> AgCl + HNO3
Mol: 0,06 0,06
\(C_{M_C}=\dfrac{0,06}{\dfrac{1}{10}.1}=0,6M\)
b, Ta có: \(\left\{{}\begin{matrix}C_{M_A}=\dfrac{n_{HCl\left(A\right)}}{\dfrac{1}{3}}=3n_{HCl\left(A\right)}\\C_{M_B}=\dfrac{n_{HCl\left(B\right)}}{\dfrac{2}{3}}=1,5n_{HCl\left(B\right)}\end{matrix}\right.\)
Mà \(C_{M_A}=4C_{M_B}\Rightarrow3n_{HCl\left(A\right)}=4.1,5n_{HCl\left(B\right)}\Leftrightarrow n_{HCl\left(A\right)}-2n_{HCl\left(B\right)}=0\)
Ta lại có: \(n_{HCl\left(A\right)}+n_{HCl\left(B\right)}=n_{HCl\left(C\right)}=1.0,6=0,6\left(mol\right)\)
Ta có hệ pt: \(\left\{{}\begin{matrix}n_{HCl\left(A\right)}-2n_{HCl\left(B\right)}=0\\n_{HCl\left(A\right)}+n_{HCl\left(B\right)}=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{HCl\left(A\right)}=0,4\left(mol\right)\\n_{HCl\left(B\right)}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_A}=\dfrac{0,4}{\dfrac{1}{3}}=1,2M\\C_{M_B}=\dfrac{0,2}{\dfrac{2}{3}}=0,3M\end{matrix}\right.\)
