a) $m_{NaCl} = 50.20\% = 10(gam)$

$m_{dd} = 50 + 20 = 70(gam)$

Suy ra : $C\%_{NaCl} = \dfrac{10}{70}.100\% = 14,29\%$

b)

$C\%_{NaCl} = \dfrac{10}{35}.100\% = 28,57\%$

\(m_{NaOH\left(35\%\right)}=100.35\%=35\left(g\right)\)

\(m_{ddNaOH\left(20\%\right)}=\dfrac{35}{20}.100=175\left(g\right)\)

⇒ m_{nước thêm vào} = 175-100 = 75(g)

_{Vnước thêm vào} = 75.1 = 75 (ml)

Câu 4:

\(n_{HCl}=4.2,75=11\left(mol\right)\)

Ta có: \(V_{ddHCl\left(2M\right)}=\dfrac{n_{HCl\left(1\right)}}{2}\left(l\right);V_{ddHCl\left(3M\right)}=\dfrac{n_{HCl\left(2\right)}}{3}\left(l\right)\)

  \(\Rightarrow V_{ddHCl\left(2M\right)}+V_{ddHCl\left(3M\right)}=\dfrac{n_{HCl\left(1\right)}}{2}+\dfrac{n_{HCl\left(2\right)}}{3}\)

  \(\Leftrightarrow4=\dfrac{3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}}{6}\Leftrightarrow3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}=24\)

Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=11\\3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2\\n_{HCl\left(2\right)}=9\end{matrix}\right.\)

  \(\Rightarrow V_{ddHCl\left(2M\right)}=\dfrac{2}{2}=1\left(l\right);V_{ddHCl\left(3M\right)}=\dfrac{9}{3}=3\left(l\right)\)

Câu 3:

\(m_{NaOH}=35.100\%=35\left(g\right)\\ m_{H_2O\left(thêm\right)}=a\left(g\right)\\ \Rightarrow\dfrac{35}{100+a}.100\%=20\%\\ \Leftrightarrow a=75\left(g\right)\\ \Rightarrow V_{H_2O\left(thêm\right)}=\dfrac{75}{1}=75\left(ml\right)\)

Câu 4:

\(Đặt:V_{ddHCl\left(pha.thêm2M\right)}=a\left(l\right)\\ V_{ddHCl\left(pha.thêm3M\right)}=b\left(l\right)\\ \Rightarrow\dfrac{2a+3b}{a+b}=2,75\left(1\right)\\ Mà:a+b=4\left(2\right)\\ Từ\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=1\left(l\right)\\b=3\left(l\right)\end{matrix}\right.\)

Vậy cần pha 1 lít dung dịch HCl 2M vào 3 lít dung dịch HCl 3M để thu được 4 lít dung dịch HCl 2,75M 

Chúc em học tốt!

Câu 2 : 

$n_{HCl} = 0,2.1 + 0,3.1,5 = 0,65(mol)$
$V_{dd} = 0,2 + 0,3 = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,65}{0,5} =1,3M$

Câu 3 : 

Gọi $m_{H_2O\ cần\ thêm} =a (gam)$

Sau khi thêm : 

$m_{NaOH} = 100.35\% = 35(gam)$
$m_{dd} = 100 + a(gam)$

Suy ra:  $\dfrac{35}{100 + a}.100\% = 20\%$

Suy ra: a = 75(gam)

Câu 4 : 

Gọi $V_{dd\ HCl\ 2M} =a (lít) ; V_{dd\ HCl\ 3M} = b(lít)$

Ta có : 

$a + b = 4$
$2a + 3b = 4.2,75$

Suy ra a = 1(lít) ; b = 3(lít)

a, Gọi \(m_{NaCl\left(thêm\right)}=a\left(g\right)\)

\(m_{NaCl\left(bđ\right)}=5\%.100=5\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{5+a}{100+a}.100\%=5,5\%\\ \Leftrightarrow a=0,53\left(g\right)\)

b, \(m_{NaCl}=58,5.5,5\%=3,2175\left(g\right)\\ n_{NaCl}=\dfrac{3,2175}{58,5}=0,055\left(mol\right)\)

PTHH: NaCl + AgNO₃ ---> AgCl↓ + NaNO₃

          0,055-->0,055------>0,055---->0,055

\(m_{AgCl}=0,055.143,5=7,8925\left(g\right)\\ m_{ddY}=58,5+200-7,8925=250,6075\left(g\right)\\ \Rightarrow C\%_{NaNO_3}=\dfrac{0,055.85}{250,6075}.100\%=1,87\%\)

pH=12=>pOH=2 =>nOH^- = 10^-2*0,2=2*10^{-3 mol}
  OH^-   +   H⁺ => H2O
0,002    0,002
sau pư, pH=1 => số mol H+ dư = 10^-1*(0,2+0,1)=0,03mol
vậy tổng số mol H+ thêm vào= 0,03+0,002=0,032mol => n_H2SO4=0,032/2=0,016mol
=>Vdd=0,016/0,6=0,0267l =26,7ml