Giúp mình với ạ aaaaaaaaa

`@` `NaCl:`

`m_(dd) = 17,55 + 202,45 = 220` `(g)`

`V_(dd) = 220/(1,1)= 200` `(mL)` `=` `0,2L`

`C% = (17,55)/(220) . 100% = 7,98%`

`n = (17,55)/(58,5) = 0,3` `(` mol `)`

`C_M = (0,3)/(0,2) = 1,5M`

`@` `Ca(OH)_2:`

`m_(dd) = 1 . 200 = 200` `(g)`

`m_(H_2O) = 200 - 14,8 = 185,2` `(g)`

`C% = (14,8)/(200) . 100% = 7,4%`

`n = (14,8)/74 = 0,2` `(` mol `)`

Đổi `200mL = 0,2 L`

`C_M = (0,2)/(0,2) = 1M`

`@` `BaCl_2:` 

`m_(ct) = (20,8 . 300)/100 = 62,4` `(g)`

`m_(H_2O) =` `300 - 62,4 = 237,6` `(g)`

`V_(dd) = 300/(1,2) = 250` `(mL)` `=` `0,25L`

`n = (62,4)/208 = 0,3` `(` mol `)`

`C_M = (0,3)/(0,25) = 1,2M`

`@` `KOH:`

`m_(dd) = 300.1,04 = 312` `(g)`

Đổi `300mL = 0,3L`

`n = 0,3 . 2,5 = 0,75` `(` mol `)`

`m_(ct) = 0,75 . 56 = 42` `(g)`

`m_(H_2O) = 312 - 42 = 270` `(g)`

`C% = 42/312 . 100% = 13,46%`

`@` `CuSO_4:`

`m_(dd) = (55,2 .100)/16 = 345` `(g)`

`m_(H_2O) = 345 - 55,2 = 289,8` `(g)`

`V_(dd) = 345/(1,15) = 300` `(mL)` `=` `0,3L`

`n = (55,2)/160 = 0,345` `(` mol `)`

`C_M = (0,345)/(0,3) = 1,15M`

$#tr$

Đáp án+Giải thích các bước giải:

 \begin{array}{|c|c|c|}\hline \text{}&\text{NaCl}&\text{$Ca(OH)_2$}&\text{$BaCl_2$}&\text{KOH}&\text{$CuSO_4$}\\\hline \text{$m_{C t}$}&\text{17,55 gam}&\text{14,8 gam}&\text{62,4 gam}&\text{42 gam}&\text{55,2 gam}\\\hline \text{$m_{H_2O}$}&\text{202,45 gam}&\text{185,2 gam}&\text{237,6 gam}&\text{270 gam}&\text{289,8 gam}\\\hline \text{$m_{dd}$}&\text{220 gam}&\text{200 gam}&\text{300 gam}&\text{312 gam}&\text{345 gam}\\\hline \text{$V_{dd}$}&\text{200 ml}&\text{200 ml}&\text{250 ml}&\text{300 ml}&\text{300 ml}\\\hline \text{$D_{dd}$}&\text{1,1}&\text{1}&\text{1,2}&\text{1,04}&\text{1,15}\\\hline \text{C%}&\text{7,98%}&\text{7,4%}&\text{20,8%}&\text{13,5%}&\text{16%}\\\hline \text{mol}&\text{0,3}&\text{0,2}&\text{0,3}&\text{0,75}&\text{0,345}\\\hline \text{$C_M$}&\text{1,5}&\text{1}&\text{1,2}&\text{2,5}&\text{1,15}\\\hline\end{array}