a)

Gọi : \(\left\{{}\begin{matrix}n_{NaCl}=a\left(mol\right)\\n_{KI}=b\left(mol\right)\end{matrix}\right.\)

NaCl + AgNO₃ → AgCl + NaNO₃

a..............a...............a..............................(mol)

KI + AgNO₃→ AgI + KNO₃

b.......b..............b..................................(mol)

Ta có :

\(n_{AgNO_3} = a + b = 0,25.2 = 0,5(mol)\)

\(m_{kết\ tủa} = 143,5a + 235b = 103,775\)(gam)

Suy ra : a = 0,15 ; b = 0,35

Vậy :

\(C_{M_{NaCl}} = \dfrac{0,15}{0,4} = 0,375M\\ C_{M_{KI}} = \dfrac{0,35}{0,4} = 0,875M\)

b)

Sau phản ứng, dung dịch gồm : \(\left\{{}\begin{matrix}NaNO_3:0,15\left(mol\right)\\KNO_3:0,35\left(mol\right)\end{matrix}\right.\)

Suy ra : 

\(m_{NaNO_3} = 0,15.85 = 12,75(gam)\\ m_{KNO_3} = 0,35.101 = 35,35(gam)\)

Số gam chất tan cần dùng để pha chế các dung dịch:

a) nNaCl = CM .V = 2,5.0,9 = 2,25 (mol)

→ mNaCl = 2,25.(23 + 35,5) = 131,625 (g)

b) 

c) nMgSO4 = 0,1.0,25 = 0,025 (mol)

→ mMgSO4 = 0,025.(24 + 64 + 32) = 3 (g)

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

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

1 - C

câu 2:

m_ct1 = \(\dfrac{150.2}{100}\) = 3 g = m_ct2

ta có:

20% = \(\dfrac{3}{m_{d2}}\).100% = 15 g

=>m_{d2 lấy ra là} = 150 - 15 = 135 g

a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

\(m_{NaOH}=0,2.40=8\left(g\right)\)

b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

\(c,C\%=\dfrac{6}{200}.100\%=3\%\)

\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)

a) nNaCl = 1.0,5 = 0,5 (mol) → mNaCl = 0,5.(23 +35,5) = 29,25 (g)

b) nKNO3 = 2.0,5 = 1 (mol) → mKNO3 = 1.101 = 101 (g)

c) nCaCl2 = 0,1.0,25 = 0,025 (mol) → mCaCl2 = 0,025(40 + 71) = 2,775 (g)

d) nNa2SO4 = 0,3.2 = 0,6 (mol) → mNa2SO4 = 0,6.142 = 85,2 (g)

Có lẽ bạn hiểu nhầm M (mol/l) với mol rồi :)

Sửa hết mol ---> M nha

\(a,n_{NaCl}=2,5.0,9=2,25\left(mol\right)\\ \rightarrow m_{NaCl}=2,25.58,5=131,625\left(g\right)\\ b,m_{MgCl_2}=\dfrac{4.50}{100}=2\left(g\right)\\ c,Đổi:250ml=0,25l\\ \rightarrow n_{MgSO_4}=0,1.0,25=0,025\left(mol\right)\\ \rightarrow m_{MgSO_4}=0,025.120=3\left(g\right)\\ d,m_{NaOH}=\dfrac{40.20}{100}=8\left(g\right)\)