\(a) m_{dd} = \dfrac{6}{15\%} = 40(gam)\\ b) V_{dd} = \dfrac{m_{dd}}{D} = \dfrac{40}{1,15} = 34,78(ml)\\ c)n_{CuSO_4} = \dfrac{6}{160} = 0,0375(mol)\\ C_{M_{CuSO_4}} = \dfrac{0,0375}{0,03478}=1,078M\)

1)

\(m_{ddCuSO_4\left(bd\right)}=1,6.25=40\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{11,25}{250}=0,045\left(mol\right)\)

=> \(n_{CuSO_4}=0,045\left(mol\right)\)

\(C_M=\dfrac{0,045}{0,025}=1,8M\)

\(C\%=\dfrac{0,045.160}{40}.100\%=18\%\)

b)

\(m_{CuSO_4}=\dfrac{200.18}{100}=36\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{5,634}{250}=0,022536\left(mol\right)\)

n_{CuSO4 (tách ra)} = 0,022536 (mol)

=> \(m_{CuSO_4\left(dd.ở.t^o\right)}=36-0,022536.160=32,39424\left(g\right)\)

\(m_{H_2O\left(bd\right)}=200-36=164\left(g\right)\)

n_{H2O (tách ra)} = 0,022536.5 = 0,11268 (mol)

=> \(m_{H_2O\left(dd.ở.t^o\right)}=164-0,11268.18=161,97176\left(g\right)\)

\(S_{t^oC}=\dfrac{32,39424}{161,97176}.100=20\left(g\right)\)

PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)

b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)

c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\) 

a) Ở 80 độ C, 100g H2O hòa tan được 40g CuSO4. 

mdd = D.V = 1,12.100 = 112 gam. ---> C% = 40/112 = 35,71%; CM = 40/160/0,1 = 2,5M.

b) m = C%.mdd = 0,2.(100+m) ---> m = 20/4 = 5 gam ---> Độ tan là 5 g.

c) mdd = 700.5/40 = 87,5 gam.

ko bt

chắc vt lộn ở đâu đó  mà tik tui đi

uk

Khối lượng CuSO₄ có trong m gam tinh thể : \(\frac{160}{250}\)m = 0,64(g)

Khối lượng CuSO₄ trong V ml dung dịch CuSO₄ c% ((khối lượng riêng bằng d g/ml) là : \(\frac{V.d.c}{100}\) = 0,01 V.d.c (g)

Khối lượng dung dịch X bằngv : m+V.d (g)

Nồng độ phần trăm của dung dịch X:

\(\frac{0,64m+0,01V.d.c}{m+V.d}.100\%=\frac{64m+V.d.c}{m+V.d}\left(\%\right)\)

\(m_{dd_{HCl\left(10\%\right)}}=150\cdot1.206=180.9\left(g\right)\)

\(n_{HCl}=\dfrac{180.9\cdot10\%}{36.5}\approx0.5\left(mol\right)\)

\(n_{HCl\left(2M\right)}=0.25\cdot2=0.5\left(mol\right)\)

\(n_{HCl}=0.5+0.5=1\left(mol\right)\)

\(V_{dd_{HCl}}=150+250=400\left(ml\right)=0.4\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{1}{0.4}=2.5\left(M\right)\)

Ta có: \(m_{ddCuSO_4}=\dfrac{3}{15\%}=20\left(g\right)\)

\(V_{ddCuSO_4}=\dfrac{20}{1,15}\approx17,39\left(ml\right)\)

Ta có: \(n_{CuSO_4}=\dfrac{3}{160}=0,01875\left(mol\right)\)

\(\Rightarrow C_{M_{CuSO_4}}=\dfrac{0,01875}{0,01739}\approx1,08M\)

Bạn tham khảo nhé!

\(n_{CuSO_4}=\dfrac{10}{160}=0,0625\left(mol\right)\\ C_{MddCuSO_4}=\dfrac{0,0625}{0,2}=0,3125\left(M\right)\\ m_{ddCuSO_4}=200.1,26=252\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{10}{252}.100\%\approx3,968\%\)