a) Giả sử lấy 100 gam dung dịch 

Sau khi thêm nước ta được dung dịch 10%

=>C%= \(\dfrac{30}{100+m_{H_2O}}=10\%\)

=> mH2O = 200 (g)

Vậy tỉ lệ : \(\dfrac{m_{H2O}}{m_{ddNaOH30\%}}=\dfrac{200}{100}=2\)

b) Sau khi cô cạn :

\(m_{dd}=\dfrac{m_{NaOH}}{50\%}=\dfrac{30}{50\%}=60\left(g\right)\)

Ta có : \(\dfrac{m_{dd}}{m_{ddNaOH30\%}}=\dfrac{60}{100}=0,6\)

Vậy cần cô cạn dung dịch giảm 0,6 lần thì được dung dịch 50% 

Câu 2:

Gọi khối lượng của ddNaCl 2% và 10% lần lượt là m₁ và m₂

\(m_{NaCl\left(1\right)}=\dfrac{2}{100}m_1;m_{NaCl\left(2\right)}=\dfrac{10}{100}m_2\)

\(\Rightarrow m_{NaCl\left(3\right)}=\dfrac{2}{100}m_1+\dfrac{10}{100}m_2=0,02m_1+0,1m_2\)   (1)

\(m_{ddNaCl\left(3\right)}=m_1+m_2\)

Ta có: \(m_{NaCl\left(3\right)}=\dfrac{8.\left(m_1+m_2\right)}{100}=0,08\left(m_1+m_2\right)\)    (2)

Từ (1)(2)

\(\Rightarrow0,02m_1+0,1m_2=0,08m_1+0,08m_2\)

\(\Leftrightarrow0,02m_2=0,06m_1\Rightarrow\dfrac{m_1}{m_2}=\dfrac{0,02}{0,06}=\dfrac{1}{3}\)

Câu 1:

\(n_{KOH}=2,4.160=384\left(mol\right)\)

\(V_{ddKOH2M}=\dfrac{384}{2}=192\left(l\right)\)

⇒ V_{nước thêm vào} = 192-160 = 32 (l)

Câu 5:

\(Đặt:V_{H_2O}=a\left(l\right)\left(a>0\right)\\ n_{KOH}=160.2,4=384\left(mol\right)\\ Vì:C_{MddKOH\left(cuối\right)}=2\left(M\right)\\ \Leftrightarrow\dfrac{384}{160+a}=2\\ \Leftrightarrow a=32\left(lít\right)\)

Vậy cần thêm 32 lít H₂O

Câu 7:

\(m_{NaOH}=200.35\%=70\left(g\right)\)

Ta có:\(m_{ddNaOH\left(20\%\right)}=\dfrac{m_{NaOH\left(20\%\right)}.100}{20};m_{ddNaOH\left(40\%\right)}=\dfrac{m_{NaOH\left(40\%\right)}.100}{40}\)

  \(\Rightarrow m_{ddNaOH\left(20\%\right)}+_{ddNaOH\left(40\%\right)}=\dfrac{m_{NaOH\left(20\%\right)}.100}{20}+\dfrac{m_{NaOH\left(40\%\right)}.100}{40}\)

   \(\Leftrightarrow400=10m_{NaOH\left(20\%\right)}+5m_{NaOH\left(40\%\right)}\)

Ta có: \(\left\{{}\begin{matrix}m_{NaOH\left(20\%\right)}+m_{NaOH\left(40\%\right)}=70\\10m_{NaOH\left(20\%\right)}+5m_{NaOH\left(40\%\right)}=400\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m_{NaOH\left(20\%\right)}=10\\m_{NaOH\left(40\%\right)}=60\end{matrix}\right.\)

\(\Rightarrow m_{ddNaOH\left(20\%\right)}=\dfrac{10.100}{20}=50\left(g\right);m_{ddNaOH\left(40\%\right)}=\dfrac{60.100}{40}=150\left(g\right)\)

\(a.\\ m_{KOH\left(A\right)}=150\cdot5\%=7,5g\\ m_{ddKOH12\%}=a\left(g\right)\\ \Rightarrow:\dfrac{7,5+12\%\cdot a}{a+150}=\dfrac{10}{100}\\ a=375\left(g\right)\)

\(b.\\ m_{NaCl\left(40^0C\right)}=1800\cdot30\%=540\left(g\right)\\ m_{H_2O\left(40^0C\right)}=1800-540=1260\left(g\right)\\ S_{20^0C}=\dfrac{540-m_{NaCl\left(kt\right)}}{1260-m_{NaCl\left(KT\right)}}=\dfrac{36}{100}\\ m_{NaCl\left(KT\right)}=135\left(g\right)\)

a. Ta có: \(C_{\%_{NaCl}}=\dfrac{m_{NaCl}}{200}.100\%=4\%\)

=> m_NaCl = 8(g)

b. Ta có: \(200=8+m_{H_2O}\)

=> \(m_{H_2O}=192\left(g\right)\)

a, mNaCl = 5% . 120 = 6 (g)

mH2O = 120 - 6 = 114 (g)

b, C%NaCl = 2 . 58,5 : 1,05 : 10 = 11,14%

nNaCl = 2 . 0,5 = 1 (mol)

mNaCl = 1 . 58,5 = 58,5 (g)

mdd = 58,5/11,14% = 525 (g)

mH2O = 525 - 58,5 = 466,5 (g)

\(a.m_{ddNaCl}=\dfrac{15}{5}\cdot100=300g\\ b.m_{nước}+m_{muối}=m_{dd,muối}\\ \Rightarrow m_{nước}=m_{dd,muối}-m_{muối}\\ =300-15\\ =285g\)